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■ ri L ly c* I m' ■ ly . .iv ■^■■',;>V ^'•- ^ ^ ^^2^«a^ 7, /f O^, r iv„^ 3 V r iui£. f J ^ ( '■"^rnr* h^ 3 - .UUj {pL{PfjL. / 3 3 Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementaryprinci01duboiala ^ / -^A^ ■V JOHN WILEY & SONS PUBLISH, By the same Author, THE ELEMENTARY PRINCIPLES OP MECHANICS. By Prof. A. J, Du Bois. Designed as a text-book for technical schools. Three Volumes, 8vo, cloth. Vol. I— Kinematics, $3.50. Vol. II— Statics, $4.00. Vol. Ill— Kinetics. $3 50 THE STRESSES IN FRAMED STRUCTURES. The present edition of this well-known work appears in a new form, greatly reduced in size and weiglit, rewritten and reset and printed from new plates. It contains the latest practice and much new matter, never heretofore published. Swing Bridges, the Braced Arch, and the Suspension System receive an entirely new treatment. New chapters are added upon Erection by John Sterling Deans, C.E., and High-Build- ing Construction, by Wm W. Crehore, C.E. Illustrated with hundreds of cuts and 35 full-page and 14 folding plates. By Prof. A. Jay Du Bois. Eleventh revised edition .' 1 vol., 4to, cloth, 10 00 HYDRAULICS AND HYDRAULIC MOTORS. With numerous practical examples for the calculation and construction of Water Wheels, including Breast, Undershot, Back-pitcli, Overshot Wheels, etc. , as well as a special discussion of the various forms of Turbines, translated from the fourth edition of Weisbach's Mechanics. By Prof. A. J. Du Bois. Profusely illustrated. Second edition 8vo, cloth, 5 00 THEORY OF THE STEAM ENGINE. Translated from the fourth edition of Weisbach's Mechanics, by Prof. A. J. Du Bois. Containing notes giving practical examples of Stationary, Marine, and liocomotive Engines, 6ho\ving American practice. By R. H. Buel. Numerous illustrations. 8vo, cloth, 5 00 THERMO-DYNAMICS, THE PRINCIPLES OF. With Special Applica- tions to Hot Air, Gas, and Steam Engines. By Robert RSntgen. With additions from Profs. Verdet, Zeuner, and Pernolet. Translated, revised, and enlarged by Prof. A. J. Du Bois, of SheflBeld Scientific School. 670 pages 8vo, cloth, r> 00 THE CALCULATIONS OF STRENGTH AND DIMENSIONS OF IRON AND STEEL CONSTRUCTIONS. With reference to the latest ex- periments. By Prof. J. J. Weyrauch, Polytechnic Institute of Stuttgart. Translated by A. J. Du Bois. With Plates 8vo, cloth, 150 %* Mailed and Prepaid on the receipt of the Priee. CATA.-L.OOTJEJi< A.lVr> C;iR<:;TJLA.FlJS OllA^TIS. THE ELEMENTARY PEINCIPLES OF MECHANICS. Vol. I. KINEMATICS. BY A. JAY DU BOIS, C.E., Ph.D., Piofessor of Civil Engineering in the Sheffield Scientific School of Yale University f Author of '^Elements of Graphical Statics," " The Strains in Fiamed Structures,'^ etc. FIRST EDITION, FIRST THOUSAND. JOHIV S. PRELL CiiAl & Mechanical Engineer. SAN FRANCISCO, CAL, NEW YORK: JOHN WILEY & SONS. London: CHAPMAN & HALL, Limited. 1899. Copyright, 1894, BY A. JAY DU BOIS. BOBBRT ORUUHOND, KLKCTHOTTPBR AND PRINTER, NEW YORK. Eiigineeruig Library Qh V.I NOTE. The large type by itself constitutes an abridged course. Articles in small type are for advanced students. Articles containing applications of the Calculus are enclosed in brackets. MECHANICS. TABLE OF CONTENTS. Vol. I. KINEMATICS. INTRODUCTIOK CHAPTER I. MJS.A8UREMEN1. PAGKS Physical Science. Definition of Kinematics, Dynamics, Statics, Kinetics, Mechanics. Measurement. Statement of a Quantity. Unit. De- rived Unit. Dimensions of a Derived Unit. Meaning of Per. Homogeneous Equations. Unit of Time. Unit of Length. Unit of Mass. Standard Units. Unit of Angle. Curvature. Tables of Measures 1 -9 CHAPTER II. POSITION. TERMS AND DEFINITIONS. Point of Reference. Position of a Point. Polar Co-ordinates. Car- tesian Co-ordinates. Dimensions of Space. Configuration. Rigid System. Rest. Motion. Path of a Point. Motion of Translation. Motion of Rotation. Combined Translation and Rotation. Material Point or Particle. Material System 10-14 GENERAL PRINCIPLES OF KINEMATICS. CHAPTER I. SPEED. Mean and Instantaneous Speed. Dimensions of Unit of Speed. Numeric Equations of Speed. Sign of Speed. Speed a Scalar Quantity. Homogeneous Equations. Examples 15-23 VI CONTENTS. CHAPTER II. BATE OF CHANGE OF SPEED. PAGES Mean and Instantaneous Rate of Change of Speed. Dimensions of Unit of Rate of Change of Speed. Numeric Equations of Rate of Change of Speed. Sign of Rate of Change of Speed. Kate of Change of Speed a Scalar Quantity. Examples. Equations of Motion of a Point under Different Rates of Change of Speed. Graphic Representation of Rate of Change of Speed. Examples . . 24-33 CHAPTER III. DISPLACEMENT. Line Representative of Displacement. Relative Displacement. Polygon and Triangle of Displacements. Resolution and Composition of Displacements. Analytical Determination of the Resultant for Concurring Displacements. Examples 34-41 CHAPTER IV. VELOCITY. Mean and Instantaneous Velocity. Unit of Velocity. Uniform and Variable Velocity. Velocity a Vector Quantity. Line Represen- tative of Velocity. Polygon and Triangle of Velocities. Compo- sition and Resolution of Velocities. Sign of Components of Ve- locity. Analytical Determination of the Resultant for Concurring Velocities. Examples 42-47 CHAPTER V. ACCELERATION. Mean and Instantaneous Acceleration. Unit of Acceleration. Uniform and Variable Acceleration. Acceleration a Vector Quantity. Line Representative of Acceleration. Polygon and Triangle of Accelera- tions. Composition and Resolution of Accelerations. Sign of Components of Acceleration. Analytical Determination of the Re- sultant for Concurring Accelerations. Equations of Motion of a Point under Different Accelerations. The Hodograph. Tangential and Normal Acceleration. Examples 48-59 CHAPTER VL MOMENT OF DISPLACEMENT, VELOCITY OR ACCELERATION. Moment about a Given Point or Axis. Line Representative of Moment of Displacement, Velocity or Acceleration. Composition and Reso- lution of Moments. Sign of Components of Moments. Algebraic Sum of Moments of Components equal to Moment of the Resultant Analytical Determination of Resultant Velocity and Moment fo'- Concurring Velocities. Examples 60-70 VOL. I. — KINEMATICS. VU CHAPTER VII. ANGULAR SPEED. PAGES Angular Revolution of a Point. Mean and Instantaneous Angular Speed. Numeric Equations of Angular Speed. Sign of Angular Speed. Rate of Change of Angular Speed. Numeric Equations of Rate of Change of Angular Speed. Sign of Rate of Change of Angular Speed. Equations of Motion of a Point under Different Rates of Change of Angular Speed. Angular Speed in Terms of Linear Speed. Rate of Change of Angular Speed in Terms of Tangential Linear. Moment of Linear Velocity in Terms of Angular Speed. Moment of Linear Tangential in Terms of Angular Speed. Normal Linear in 1'erms of Angular Speed. Motion in a Circle. Graphic Representation of Rate of Change of Angular Speed. Examples 71-80 CHAPTER VIII. DIFFERENTIAL EQUATIONS OF MOTION OF A POINT. Free Motion of a Point, Rectangular Co-ordinates. Application of Formulas. Differential Polar Equations for Motion of a Point in a Plane. General Polar Equations for Motion of a Point in a Plane, Acceleration Central. Differential Equations for Constrained Motion of a Point in a Plane 81-90 KINEMATICS OF A POINT. CHAPTER I. RECTILINEAR MOTION. CENTRAL ACCELERATION. Uniform Acceleration, Motion Rectilinear. Value of g. Formulas for a Body Vertically Projected. Application of Calculus. Examples. Acceleration Inversely as the Square of the Distance from a Fixed Point. Application of Calculus 91-102 CHAPTER IL RECTILINEAR SIMPLE HARMONIC MOTION. RESISTING MEDIUM. Simple Harmonic Motion. Simple Harmonic Motion in a Straight Line. Amplitude, Epoch, Period, Phase. Application of Calculus. Examples. Body Falling iu a Resisting Medium. Body Projected Upwards in a Resisting Medium. Values of d, ^andc. Examples. 103-116 CHAPTER III. TRANSLATION IN A CURVED PATH— PROJECTILES. Curved Path. Uniform Acceleration Inclined to Direction of Motion. Theory of Projectiles. Application of Calculus. Examples. Motion of Projectile in a Resisting Medium 117-129 Viii CONTENTS. CHAPTER IV TRANSLATION IN A CURVED PATH— HARMONIC AND PLANETARY MOTION. PAGES Central Acceleration. Cases of Central Acceleration. Cases of Har- monic Motion. Resolution and Composition of Harmonic Motion. Graphic Representation of Harmonic Motion. Blackburn's Pendu- lum. Application of Calculus. Planetary Motion. Path for Planetary Motion. Kepler's Laws. Verification by Application to the Moon. Application of Calculus. Value of a' for Planetary Motion. Examples 130-150 CHAPTER V. CONSTRAINED MOTION OF A POINT. Motion on an Inclined Plane. Uniform Acceleration. Examples. Mo- tion in a Curved Path, Uniform Acceleration. Motion in a Circle, Uniform Acceleration. Examples. Motion in a Cycloid, Uniform Acceleration. Application of Calculus. The Brachistochrone. . . . 150-160 Miscellaneous Problems 161-168 KINEMATICS OF A EIGID SYSTEM. CHAPTER I. RIGID SYSTEM WITH ONE POINT FIXED-ROTATION. Rotation. Angular Displacement of a Rigid System. Linear Dis- placement in Terms of Angular. Line Representative of Angular Displacement of a Rigid System. Composition and Resolution of Angular Displacements. Sign of Components of Angular Dis- placement. Axis of Rotation. Mean and Instantaneous Angular Velocity of a Rigid System. Instantaneous Axis of Rotation. Unit of Angular Velocity. Unifonn and Variable Angular Ve- locity. Mean and Instantaneous Angular Acceleration of a Rigid System. Instantaneous Axis of Angular Acceleration. Com- position and Resolution of Angular Velocities and Accelerations. Unit of Angular Acceleration. Relations between Angular and Linear Velocity and Acceleration. Equations of Motion of a Rotating Rigid System under Different Angular Accelerations. Moment of Angular Displacement, Velocity or Acceleration. Line Representatives of Angular Displacement, Velocity or Accelera- tion. Condition for Rotation only. General Analytical Deter- mination of Resultant Angular Displacement, Velocity or Ac- celeration for any Number of Concurring Components. Result- ant of Two Concurring Component Angular Displacements, Velocities or Accelerations. Examples 169-185 CHAPTER II. TRANSLATION AND ROTATION. Moment of a Couple. Composition and Resolution of Translation and Angular Displacement. Displacement of a Rigid System. VOL. I. — KINEMATICS. IX Composition and Resolution of Translation and Angular Velocity or Acceleration. Central Axis. Screw Motion. Centre of Paral- lel Angular Velocities. Rotation and Rectilinear Translation Combined. Combined Parallel Rotations, One Axis Fixed. Rotation about Intersecting Axes, One Axis Fixed. Analytical Determination of Resultant Angular Velocity and Velocity of Translation for a Rigid System with any Number of Non-concur- ring Angular Velocities. Parallel Velocities. Components of Motion of a Rigid System. Composition and Resolution of Screws. Examples. Relative Motion of a Body. Acceleration of Relative Motion 186-218 CHAPTER III. Blow ROTATING SYSTEM. General Analytical Relations for a Point of a Rigid Rotating System. Euler's Geometric Equations. Examples 219-225 Index .,.., 227-231 INTRODUCTIOK CHAPTEB L MEASUEEMENT. Physical Science. — We live in a world of matter, space and time. We do not know what these are in themselves and we cannot ex- plain or define any one of them in terms of the others. Thus we recognize matter in certain states which we call solid, liquid or gaseous. We distinguish also different kinds of matter, such as iron, wood, glass, water, air, etc., which we call substances. We also recognize limited portions of matter of definite shape and volume, such as a pebble, a rain-drop, a planet, etc., which we call material bodies. But what matter is in itself we do not know. We also recognize matter as occupying space and we note suc- cessive events as occupying time. But what space and time are in themselves we do not know. We also recognize force as causing change of motion of matter. But what force is in itself we do not know. Yet although we thus know nothing of matter, space, time and force in themselves, we can and do investigate them in their measur- able relations, and such investigation is the object of all physical science. Mechanics — Kinematics and Dynamics — Statics and Kinetics. — That branch of physical science which treats of the measurable relations of space alone is called geometry. That which deals with the measurable relations of space and time only, that is with pure motion, is called kinematics {Kivrffi >, motion). To the ideas of geometry it adds the idea of motion. That which deals with the measurable relations of space, time and matter, involved in the study of the motion of material bodies under the action of force, is called dynamics {8vvai.ni^ force). To the ideas of kinematics it adds the idea of force. We divide dynamics into two parts : statics, which treats of material bodies at rest under the action of force, and kinetics, which treats of material bodies in motion under the action of force. Statics is thus a special case of dynamics which it is con- venient to consider separately. In the study of machines, or of moving bodies generally, under the action of force, we have to make use both of the principles of 2 INTRODUCTION". [CHAP. I. kinematics and of dynamics. The term mechanics is therefore used to include the general principles of both kinematics and dynamics, while their special application to machines is called applied me- chanics or mechanism. We treat in this work of mechanics as thus defined, or the general principles of kinematics and dynamics. We have then as the scheme of the present work : C Vol. I, Kinematics ; Mechanics: -s Vol. II, Statics; / -n^^ ««„;«» (Vol. m; Kinetics; [ I>y^amics. Measurement. — Since then we have to do in all that follows with the measurable relations of force, matter, space and time, the sub- ject of the measurement of these quantities should first engage our attention. Unit. — In order to measure any quantity whatever, we must always compare its magnitude with the magnitude of another quantity of the same kind. The quantity thus taken as a standard of comparison is called the unit of measurement. Thus the unit of length must itself be some specified length, as, for instance, one foot, one yard, one centimeter or one meter. The unit of time must be a specified time, as one second. The unit of mass must be a specified mass, as one pound or one gram or one kilogram. The units of mass, length and time are called fundamental units, because not derived from any others. Statement of a Quantity. — The complete statement of a quantity requires, therefore, a statement of the unit adopted and also a state- ment of the result of comparison of the magnitude of the quantity with the magnitude of the unit. The result of this comparison is always a ratio between the magnitudes of two quantities of the same kind and is, therefore, always an abstract number. This ratio or abstract number is called the numeric. Thus we say 3 feet, 4 seconds, 5 pounds. In each of these cases we state both the unit and the numeric, or ratio of the magnitude of the quantity to that of the unit. Thus 3 feet denotes a quantity whose magnitude is three times the magnitude of one foot. So for any quantity. In general, if L stands for any length and [L] stands for the unit of length, we have L = l[L], or the length equals I times the unit of length. Here I is the numeric and is an abstract number. Again, if Tis a certain interval of time, and [T] stands for the unit of time, we have T= t[T], or the time equals t times the unit of time. Here t is the numeric and is an abstract number. So also if ilf is a certain mass and [M] stands for the unit of mass, we have M= m[M], or the mass equals m times the unit of mass. Here m is the numeric and is an abstract number. Derived Unit.— A unit of one kind which is derived by reference . to a unit of another kind is called a derived unit. Thus the unit of area may be taken as a square whose side is one unit of length, or one square foot. The unit of volume may be taken as a cube whose edge is the unit of length, or one cubic foot. The unit of speed may be taken as one unit of length per unit of tim^, or one foot per second. Such units are derived units, while the units of mass, space and time, not being thus derived from any others, are fundamental units. CHAP. I.] MEASUREMENT — HOMOGENEOUS EQUATIONS. 3 Dimensions of a Derived Unit.— A statement of the mode m which the magnitude of a derived unit varies with the magnitudes of the fundamental units which compose it is a statement of the dimensions of the derived unit. Thus let [A] denote the unit of area and \L] the unit of length. Then if A — a[A\ is the area of a square whose side is L = l[L], where a and I are abstract numbers, we shall have a[A] — riLf. Now we shall have the numeric equation a — l'\ or the number of units of area equals the square of the number of units of length, provided we have [ A] = \L]'', or the unit of area equal to the square of the imit of length. The statement [AJ = [Ly is a statement of the dimensions of the unit of area. Again, let [L] denote the unit of length and [T] denote the unit of time and [V] denote the unit of speed. Then if Z, = Z[iJ is any distance and T = t[T] is the time occupied in describing that distance, and the mean speed is F"= v[V], we have We shall then have the numeric equation v = y, or the number of units of speed is equal to the number of units of length passed over divided by the number of units of time occupied, provided we have L y J (-jry or the unit of speed equal to one unit of length per unit of time. This is a statement of the dimensions of the unit of speed. Meaning of "Per." — It will be observed that the statement [ V] = iJ is read, " the unit of speed is equal to the unit of length per unit of time," and the word per is indicated by the sign for ''divided by.'' Now we can divide the numeric I by the numeric t, because these are abstract numbers. But it would be nonsense to speak of divid- ing length by time, or a unit of length by a unit of time. We there- fore avoid such a statement by the use of the word per. If then we give to the symbol of division this new meaning, we can then treat it by the rules which apply to the old meaning, and thus avoid the invention of a new symbol by using an old one in a new sense. Wlienever, then, the tvord " per " is used, it can be replaced by the sign of division. Homogeneous Equations. — The symbols in all formulas or state- ments of the relations of quantities always stand for the numerics of these quantities, and the units are always understood though not written. Thus such an equation as u = ^ or Z = rHs a numeric equation, and the units are understood and must always be supplied in inter- preting them. When the units are thus supplied, all the terms on both sides of the equation which are combined by addition or sub- traction must always be of the same kind, whatever the system of units adopted. Such equations are called homogeneous. If any numeric equation is not thus homogeneous, it is incor- rectly stated. ■4 INTRODUCTIOJSr. [CHAP, I. It is also evident that all algebraic combinations of such homo* ;geneous equations must always produce homogeneous equations. If not, some error must have been made in the algebraic work. Error can thus often be detected in the result of an investigation without following through its successive steps, by simply inserting the omitted units, and no equation or result should be accepted, or even discussed, which does not stand this test. Thus in the equation I — vt, if we supply the omitted units, we iiave l[L\ = v"^ X t[T] = vt[L]. The equation is therefore homogeneous, since the unit of length is to be understood in both terms. Unit of Time. — The unit of time ordinarily adopted in dynamics is the second or some multiple of the second. It is the time of vibration of an isochronous pendulum which T^ibrates or beats 86400 times in a mean solar day of 24 hours, each hour containing 60 minutes and each minute 60 seconds (24 x 60 ^ 60 = 86400). The sidereal day contains 86164.09 of these mean solar seconds. Unit of Length. — The unit of length ordinarily adopted in dynamics is the foot or the meter or some multiple of these. Unit of Mass. — The unit of matter or mass ordinarily adopted in •dynamics is the pound or the kilogram. Standard Unit. — All units adopted are defined by reference to certain standard units. A standard unit, in general, should possess, so far as possible, a permanent magnitude unchanged by lapse of time and unaffected by the action of the elements or by change of place or temperature. It should be capable of exact duplication and should admit of direct and accurate comparison with other quantities of the same kind. Standard Unit of Time. — The standard unit of time is the period of the earth's rotation, or the sidereal day. This has been proved by Laplace, from the records of celestial phenomena, not to have changed by so much as one eight-millionth part of its length in the course of the last two thousand years. The length of the solar day is variable, but the mean solar day, which is the exact mean of all its different lengths, is the period already mentioned, which furnishes the second of time. It is 1.00273791 of a sidereal day. The second can therefore be defined, with reference to the stand- ard unit of time, as the time of one swing of a pendulum so ad- justed as to make 86400 oscillations in 1.00273791 of a sidereal day. Standard Units of Length. — The English standard unit of length is the length of a standard bronze bar, deposited in the Standards Department of the Board of Trade in London. Since such a bar changes in length with its temperature, the length is taken at the specified temperature of 62° Fah. The length of this bar at this temperature is the English stand- ard unit of length, and is called the standard yard. Accurate copies of this standard are distributed in various places, and from these all local standards of length are derived.* The foot is defined as one third the length of the standard yard at 62° Fah. * The English standard yard is 1 part in 17230 shorter than the U. S. copy. CHAP. I.] MEASUREMENT — UNIT OF ANGLE. 5 The French standard of length is the meter, and is the length of a bar of platinum at the temperature of melting ice, or 0° C. This bar is preserved at Paris. Its length was intended to be the ten- millionth part of a quadrant of the earth's meridian through Paris. The quadrant of the meridian through Paris is 10001472 stand- ard meters, according to Colonel Clarke's determinations of the size and figure of the earth, which are at present the most authori- tative, and thus the standard Paris meter is slightly less than the- length upon which it was founded. The material bar is therefore- the standard, just as is the case with the English standard. The relation of the meter to the meridan was intended as a means of reproduction in case of destruction of the standard, but in such case the standard would probably be reproduced from the best existing copies. This was actually the case with the original English standard, which was destroyed by fire in 1834. It had been originally de- fined as having at 62° Fah. a length of - of the length of a pendulum vibrating seconds in the latitude of London at the sea- level. But this provision for its restoration was repealed and a new standard bar was constructed from authentic copies of the old one. The English inch, or the 36th part of the length of the standard yard, is very nearly equal to the five-hundred-millionth part of the length of the earth's polar axis [-^q^^q) The utUity of the standard, however, does not depend upon any such earth relations, the only value of which is for reproduction in case of destruction — a value which, as we have seen, is practically disregarded. The ultimate standards are therefore the actual bars. Standard Units of Mass.^The English standard unit of mass is a piece of platinum deposited in the Office of the Exchequer at London and called the "Imperial Standard Pound Avoirdupois." The French standard unit of mass is a piece of platinum pre- served at Paris and called the kilogram. Unit of Angle. — There are two units of angle in use, the degree and the radian. The degree is that angle subtended at the centre of any circle by an arc equal in length to ^^ part of the circumference of that circle. It is subdivided sexagesimally into degrees, (°), minutes ('), and seconds ("). The seconds are subdivided decimally. Minutes and seconds of time are distinguished by being written mm., sec. The radian i« that angle subtended at the centre of any circle by an are equal in length to the radius. It is subdivided decimally. If then the length of any arc is s[L], or s units of length, and the length of the radius is r[L], or r units of length, and if the angle subtended at the centre is 6 radians, we have „ slL] s , 6 = -t— ^ = - , or rB =s. r{L\ r The number of radians in any angle is then found by dividing the- number of units of length in the subtending arc by the number of units of length in the radius, and this number is independent of the particular unit of length adopted, whether feet or centimeters. 6 INTKODUCTION. [CHAP. I. If the subtending arc is the entire circumference, the number of radians is — = 2*. Hence 27t radians correspond to 360 degrees, or 1 radian corresponds to — = = 57.29578 degrees — 57° 17' 44".8. Any angle expressed in radians may then be converted into 180° degrees by multiplying the number of radians by = 57.29578 degrees = 1 radian. Any angle expressed in degrees may be converted into radians by multiplying the number of degrees by -— = 0.0174533 radians 180 = 1 degree. (1) Express 12° 84' 56" in terms of radians; and 3 radians in terms of degrees. Ans. 0.2196 radians ; 171° 53' 14".424. (2) The radius of a circle is 10 feet; what is the angle subtended at the center by an arc of 3 feet f 3 Ans. -r radian, or 17° 11' 19. "44. (3) How much must a rail 30 feet long be bent in order to fit into a curve of half a mile radius f Ans. 5-^ radian, or 0° 39' 3". 92. 00 (4) Express 45 degrees in terms of radians, and 4.5 radians in terms of degrees. Ans. J radians = 0.7854 radians ; 257° 49' 51". 636. 4 (5) The avgle subtended at the centre of a circle by an arc whose length is 1.57 feet is 15° ; what is the radius f s 157r 157X180 „ „^ Ans. - = 757, , or r = — ^g = 6 ft. r 180 15n: (6) What is the sin — radians; cos — radians; cos — radians; 6 o 3 tan — radians f 4 Ans. 0.5; ^ VS; 0.5; 1. (7) Express in degrees and in radians the angle made by the hands of a clock at 35 minutes past 3 o^clock. Ans. 102.5 degrees; 1.79 radians. Unit of Conical Angle. — Let the area of any portion of the sur- face of a sphere be A[A], or A units of area, and let the square of the radius be r'^[A], or r'' units of area. If lines are drawn from the centre C of the sphere to every point of the area, they form a cone, and the angle subtended at the centre C by the area we call a conical angle. The conical angle subtended at the centre of a sphere by a por- tion of its surface whose area is equal to the square of its radius we CHAP. I.] MEASUJIEMENT — CURVATURE. 7 call a square radian. If we denote the conical angle subtended by the area A by 9 square radians, we have A[A] = — , or r^9 = A. r'iA] r The number of square radians in any conical angle is thu3 found by dividing the number of units of area in the subtending area by the number of units of area in the square of the radium, and this number is independent of the unit of area adopted. If the subtending area is the entire surface of the sphere, the number of square radians is — 5- = 4;r. Hence the surface of a r sphere subtends a conical angle of 4n: square radians. [The terms solid angle and solid radian are usually employed in place of conical angle and square radian. as defined, but as they seem in no way descriptive, we have employed the latter terms as more expressive.] Curvature. — The direction of a plane curve at any point is that of the tangent to the curve at this point. Thus the direction of the curve AB at the point A is that of the tangent AC. The change of direction be- tween any two points of a plane curve is the angle be- tween the tangents at these two points, and is called the integral curvature. Thus the angle 6, or change of direction between the tangents at A and B, is the integral curva- ture for the curve between A and B. The integral curvature for any portion of a plane curve divided by the length of that portion is the mean curvature. Thus if the length from A to -B of the curve is s[L], or s units of length, the mean curvature is ,M-- Since 6 is given in radians, the sill] unit of curvature is one radian per unit of length of arc. When we say, therefore, that the mean curvature is - , we mean — radians per unit of length of arc. The limiting value of the mean curvature when the two points are indefinitely near is called the curvature. The curvature, therefore, is the limiting rate of change of direc- tion per unit of length of arc. Its unit is one radian per unit of length of arc. Curvature of a Circle. — If the curve is a circle, the angle at the centre between the radii at A and B will be equal to the angle 9 between the tangents at A and B. We have then 9 = — radians. The naean curvature is then r 6 1 _ = — radians per unit of length of arc. s T Since this is independent of s, the curvature at every point of a circle is constant and equal to the mean curvature for any two points, viz., — radians per unit of length of arc. r 8 INTRODUCTION. [chap, r. Curvature of any Plane Curve. — For any plane curve whatever a circle can always be described whose cui-vature is the same as that of the given curve at the given point. This is the circle of curvature of the curve at that point. Its radius is the radius of curvature of the curve at that point. If then p is the radius of curvature of a curve at any given point, the curvature of that point is — radians per unit of length of arc. Since curvature then depends only upontheradius of curvature, the circle is the only curve whose curvature is constant. (1) A circle has a radius of 10 feet. What is its curvature f Ans. f\f radian per foot of arc, or 5°. 73 per foot of arc. (2) If the radius is 10 yards, what is the curvature f Ans. -^Tj radian per yard of arc, or 5°. 73 per yard of arc, or -^ radian per foot of arc, or 1°.91 per foot of arc. Dimensions of Unit of Curvature. — If C is the curvature and c the number of units of curvature, we have by definition c[C] = -fJr, where [C] is the unit of curvature, and [0] is the unit of angle, [L] s[L] the unit of length, and 6, s the number of units of angle and length. We shall always have c = — , provided we take [C] = f^, that is, 8 [L] Erovided the unit of curvature is equal to the unit of angle divided y the unit of length. This is a statement of the dimensions of the unit of curvature. The unit of curvature is then one unit of angle per unit of length of arc, as, for instance, one radian per foot of arc, or one de- gree per foot of arc. A railway curve has a length of one mile, the curvature is uniform, and the integral curvature is 30 degrees. What is the curve, the curvature, and the radius of curvature f Ans. A circle ; 0.5336 radian per mile arc ; 1.9 miles radius. Tables of Measures. — We shall deal in the course of this work with many other derived units, which will be explained as they occur. It will be useful to collect here for convenience of reference a number of such units. I. MEASURES OF SPACE. A. LENGTH. Centimeters I I I I 10 Table 1. 1 centimeter = 0.3937079 inch 1 Tr,«+o, _ j 39 37079 inches 1 meter -•) 3.2809 feet 1 kilometer 2 Inches _ { 0.62137 mile "~ (0.535987 nautical m. Table 2. 1 inch = 2.539954 centimeters 1 foot = 30.479449 1 yard = 0.91438347 meter 1 mile = 1.60935 kilometers CHAP. I.J MEASUREMENT — TABLES OF MEASURES. 9 The following are approximate : The centimeter is about | inch. The meter is about 3 ft. 3f inches. The decimeter is about 4 inches. One kilometer is about f of a mile. Distance from pole to equator is about 10000000 meters. Earth's polar radius is about 500000000 inches. B. AREA. Table 3. Table 4. 1 sq. centimeter = 0.155006 sq. inch 1 sq. inch = 6.45137 centimeters 1 sq. meter = 10.7643 sq. feet 1 sq. foot = 928.997 1 sq. hectometer, I _ o ati-ia a^^^c 1 sq. vard = 0.836097 meter or 1 hectare ) " •^•^'^^* ^^^^^ 1 acre' = 0.404672 hectare 1 sq. kilometer = 0.38611 sq. mile 1 sq. mile = 2.58989 sq. kilometers C. VOLUME. Table 5. Table 6. 1 cubic centimeter = 0.0610271 cu. inch 1 cubic inch=16.3866cubic centimeters 1 liter, or 1 cubic } = 61 .0271 cu. inches 1 cubic foot=28.31o3 liters decimeter f =1.76172 pints 1 cubic yard=0. 764513 cubic meter 1 cubic meter =35.3166 cubic feet 1 pint =0.567627 liter 1 gallon =4.54102 liters II. MEASURES OF MASS. Table 7. Table 8. 1 centigram = 0.154323 grain 1 grain = 0.064799 gram 1 _ j 15.4323 grains 1 oz. = 28.3496 grams 1 gram _ ^ 0.0353739 oz. 1 lb. or 16 oz. = 0.453593 kilogram' 1 kilogram = 2.20463 lbs. 1 ton or 2240 lbs. = 1016.05 kilos 1 gram = mass of 1 cubic centimeter of pure water at 4° C. 1 kilogram = mass ot 1 liter of pure water at 4" O. 1 gallon = 277.274 cubic inches. The gallon contains 10 lbs. of pure water at 62" F. 1 cubic foot of water contains about 1000 oz. or 62^ lbs. The pint contains 20 fluid oz. Acceleration of gravity at London = 32.182 feet-per-second per second = 980.889 centimeters-per-second per second. Average value 32^ feet-per-second per second or 980.3 centimeters-per-second per second. 1 dyne = force which will give a mass of 1 gram an acceleration of 1 centi- meter- per-second per second = about j^y weight of gram = weight of about 1 milligram. 1 poundal = force which will give a mass of 1 pound an acceleration of 1 foot-per-second per second = about weight of | oz. CHAPTER 11. POSITION. TEEMS AND DEFINITIONS. Point. — A mathematical point has neither length, breadth, nor thickness. It is therefore without dimensions and indicates posi- tion only. Point of Reference. — When we speak of a point as having position, some other point or points must always be assumed, by reference to which the position is given. Such a point is a point of reference. It is also called a pole, or origin. Position then is always relative. We know nothing of "abso- lute" position. Thus tlie position of the point C is known with ^ respect to A when we know the length of the line AO and the angle BA C or the direction of the line A G. The points A and B are points of reference, by means of which C is located. Position of a Point. —The position of a point with reference to other assumed points is then known when we have sufficient data to locate it. These data give rise to two methods of location : 1st, by polar co-ordinates. 2d, by Cartesian co-ordinates, so called because first employed by Descartes. Plane Polar Co-ordinates.— The data necessary for locating a point by polar co-ordinates, when the point is situated in a given plane, consist of a distance and an angle. If the point is not in a known plane, of a distance and two angles. Thus, if the point P, in the plane of this page, is to be located, we first assume a line OA in the plane, as a line of reference. Then the position of P with reference to O is given by the angle AOP and by the distance OP. The assumed point O is called the pole; OA is the line of reference; the distance OP is^ called the radius vector, and its magnitude is usually denoted by r ; the angle AOP is the direction angle ; its magnitude is denoted by 0, and it is measured around from OA to the left. The polar co-ordinates for a point in a given plane are therefore r and 0, or a distance and an angle. These are plane polar co- ordinates. Space Polar Co-ordinates.— If the point P is not in a given plane, 10 CHAP. II.] TERMS AND DEFINITIONS. 11 we assume as before a pole O, and a reference line OA in space. Through this line we assume any plane, as the plane of this page, OABC, and let OB be the inter- section of this plane with a plane OPB, perpendicular to it and passing through OP. The location of P is then given by the length OP or the radius vector r, the angle AOB or (p, and the angle BOP or Q. The polar co-ordinates for a point not in a given plane are there- fore r, (p and 0, or a distance and two angles. These are spaxie polar co-ordinates. If O is a point on the earth's surface, and the reference line OA is a north and south line in the plane of the horizon, the angles S and would be the astronomical azimuth and altitude of the point P. Cartesian Co-ordinates. — Plane.— The data necessary for locating a point by Cartesian co-ordinates, if the point is in a known plane, consists of two distances, parallel to two assumed lines of reference in that plane, passing through the point of reference, which is called the origin. The assumed lines of reference are usually taken at right angles. Thus if the point P is known to be in the plane of this page, we assume any origin O and draw two reference lines OX and 01^ through O in this plane and at right angles. These two lines are called the axes of co-ordinates, the horizontal one the axis of x, or the x axis, the other the axis of y, or the y axis. The distance BP or OA is denoted hjx and called the abscissa of the point P. The distance APis denoted by y and called the ordinate of the point P. The abscissa x is positive to the ri^ht, negative to the left of the origin, while the ordinate y is positive when laid off above and negative when below the origin. Any point in the plane is thus located with respect to O. If a point IS in the first quadrant, its co-ordinates are + x, -f ^; if in the second quadrant, —x and -i- y; if in the third quadrant, — X and —y; if in the fourth quadrant, -f x and — y. When the point is in a known plane, the co-ordi- nates are called plane co-or- dinates. Space Co-ordinates. — If the point P is not in a known plane, we take three axes through the origin, all at right angles usually. Two of these we may denote by X and Y as before ; the third, at right angles to the plane of XY, z we call the axis of z, or the z axis. 1 •-a; r p +y +05 +2/ -y o 3 3 4 A -y 3 1^ f / / / 1 ^ ^ 1 y ; o ^' X / /A 12 INTRODUCTION. [CHAP. 11^ Thus the position of the point P is given by the distance OA = X, the distance AC = z, and the distance CP = y. These are the space coordinates of the point P. The signs prefixed to the co-ordinates indicate the quadrant in which the point is located as before. Thus, + x, + y and ± z denote a point in the first quadrant either in front of or behind the plane of XY\ — x, + y and ± z, a point in the second quadrant, either in front of or behind the plane of XY; — x, — y, ± z, and + X, — y, ± z, points in the third and fourth quadrants, either in front of or behind the plane of XY. Direction Cosines. — If we join the origin O and the point P by a line, and denote the angle of OP with the x axis by «, with the y axis by ^, and with the z axis by r, we have the relations X = OP cos a, y = OP cos /i, z = OP cos y. These cosines are called the direction cosines of OP. Since OP is the diagonal of a parallelogram, we have OP" = a;' + 2/' + 2' = OP' (cos" a + cos" /i + cos" r). Hence cos" a + cos" /3 + cos" X = 1 (I)' If, therefore, any two of these direction cosines are given, the third can always be found. Since cos 2a = 2 cos" cz — 1, cos 2/3 = 2 cos" /^ — 1, cos 2y — 2 cos" y — 1, yve have also cos 2a + COS 2/3 + cos 2r = — 1 (2) Again, since cos (a + fi) = cos a cos /? — sin « sin /S, cos {a — ft) = cos a cos /3 + sin a sin ft, we have also cos (« + ft) COS {a - /3) + cos" X = (3> Dimensions of Space. — A point in a given line is at once located by a statement of the distance of the point from either end of the line. A point in a given plane is located either by two distances or by a distance and an angle. A point in space is located either by three distances or by a distance and two angles. Hence space is said to have three dimensions, a plane surface to have two dimensions, and a line one dimension. A point has no dimensions and indicates position' only. System.— Any definite and limited assem- _ blage of points is called a system. Thus the B assemblage of points represented by A, B, C,. constitutes a system. Configuration. — The relative position or arrangement of any system or assemblage of points is called the confignration of the system. A knowledge of the configuration of a system at any instant, requires a knowledge of the relative position of every point of the system with reference to every other point at that instant. Thus the configuration at any instant of the system represented by the. points A, B, C, is known when the position at that c instant of A relative to B and C, of B relative to ^ and C, and of C relative to A and B, are known ; that is, when all the sides and angles of the triangle are known at that instant. Rigid System. — When the configuration ^ does not change, the system is a rigid system. CHAP. II.] TERMS AND DEFINITIONS. 13 Thus when the angles and sides in the triangle ABC remain un- changed, the system is rigid] Rest. — When the straight lines drawn from a point to any as- sumed points of reference do not change their length or inclination to each other, the point is at rest with reference to these points. The points of a rigid system are therefore at rest relatively to each other. The entire system may, however, be in motion with reference to some external point. Rest, then, like position, is relative only. We know nothing of " absolute" rest. Thus if the lines CA and CB in the preceding figure do not change in length and the angle ACB does not change, the point (7 is at rest with refer- ence to A and B. The system is then a rigid system, and A is at rest with reference to C and B, and B is at rest with reference to A and C. The entire system of ^, B, Cmay, however, be in motion with reference to «ome point external to the system. The stationary objects in a room are all at rest relatively to each other. The straight lines joining any three do not change their length or inclination to each other. But, as we know, all these objects partake of the motion of the earth, and -are therefore not at rest with reference to the sun. Motion. — Motion is change of position. A point moves when the straight lines joining it to the points of reference change either in length or inclination to each other. Motion, therefore, like rest and position, is always relative. Such terms as " absolute " position, "absolute " rest, and " absolute" motion have no scientific value. All our knowledge, both of posi- tion and change of position, must be essentially relative. "When a man has acquired the habit of putting words together, without troubling himself to form the thoughts which ought to correspond to them, it is easy for him to frame an antithesis between this relative knowledge and a so-called absolute knowledge, and to point out our ignorance of the absolute position of a point as an instance of the limitation of our faculties. Any one, however, who will try to imagine the state of a mind conscious of knowing the absolute position of a point will ever after be content with our relative knowl- ■edge." ("'Matter and Motion," by J. Clerk Maxwell. Pott, Young & Co. , New York, 1876.) Path of a Point. —The line joining the successive positions of a point during its motion is called its path. The path thus described by a point may be either a straight or a <;urved line, or a combination of straight and curved lines. If the path is without angles or abrupt changes of direction, it is con- tinuous. Motion of Translation. —When a rigid system moves so that every straight line in it joining every two points remains always parallel to itself, the system is said to have a motion of translation. The paths of all the points are therefore parallel at every instant and equal for any given interval of time, and the translation of the system is that of any one of its points. Motion of Rotation. — When a rigid system moves so that all its points describe arcs of circles in parallel planes about a common straight line or axis passing through the centers of the circles and perpendicular to their planes, the system is said to rotate or have a motion of rotation about that axis. Since the system is rigid, every point must describe an equal angle in the same time. 14 INTRODUCTION. [CHAP. II. A point has no dimensions and therefore cannot have motion of rotation, but only one of translation. Combined Translation and Rotation. — A rigid system may have a motion of rotation and translation at the same time. Thus, for instance, a rolling wheel has a motion of rotation about an axis through the hub at right angles to the plane of the wheel, while at the same time every point of the wheel has a motion of translation. Material Point or Particle. — We have just seen that a point can- not have motion of rotation. Kotation is possible only to systems of points. A material body so small that the distances between its points may be neglected is called a particle. When the body is not small, whatever its magnitude, if the dis- tances between its various points have no influence upon the motion considered, we call the body a material point or particle and may represent it by a point without dimensions. Thus if we are investigating the motion of the earth about the sun, so far as the motion of translation of the earth is concerned, we may regard both the earth and sun as points. But we cannot treat them as points v, hen wo wish to study their rotation. Material System. — A number of material points or particles con- stitute a material system. When we confine our attention to such a system, all relations or actions between one point of such a system and another are called internal relations or actions. Those between the whole or any part of the system and bodies not included in the system are called external relations or actions. V^' KINEMATICS. GENERAL PRINCIPLES. CHAPTER I. SPEED. Kinematics. — That branch of science which treats of the measur- able relations of time and space only, that is, of pure motion, is called kinematics. It adds to the ideas of pure geometry the idea of motion. Mean Speed of a Point, — The distance described by a moving point per unit of time is called the mean speed of the point. There- fore the number of units of distance described in a given time, divided by the number of units in that time, gives the number of units of mean linear speed. The mean speed is then the mean time-rate of motion in the path. When the mean speed varies with the interval of time it is variable. When it has the same magnitude no matter what the intei'val of time it is uniform. A point moving with uniform mean speed evidently describes equal distances in equal times. Instantaneous Speed of a Point. — The limiting value of the mean speed when the interval of time is indefinitely small is called the instantaneous speed. When the instantaneous speed at any instant is equal to the mean speed for any interval of time it is uniform. When the instan- taneous speed is variable the mean speed has different values for equal intervals of time. The term speed always signifies instantaneous speed unless otherwise specified. Dimensions of the Unit of Speed. — Let us denote any speed by V, the unit of speed by [V] and the number of units of speed by v, so that V =v[V]. Then if [L] is the unit of length and s the num- ber of units of length, or the distance passed over in the time t[T]y we have by definition ^^ t[T] o We shall always have the numeric equation v = —, if we take 15 16 KINEMATICS — GEKERAL PRINCIPLES. [CHAP. I. t^] = ^» ^^ *^® ^"^* ®^ speed equals the unit of length per unit of time. This is the statement of the dimensions of the unit of speed. The unit of speed is therefore always taken as one unit of length per unit of time, as, for instance, one foot per second. Numeric Eq^uatious of Speed. — If si denotes, then, the number of "^ units in the initial distance OA, measured along the path, of a moving point from a fixed point O in the path, taken as origin, and s the number of units in the final distance OB, measured along the path, from the same origin O, and t the number of units in the interval of time in describing the distance AB = s — Si, we have for the mean speed the numeric equation ^ _ / ^ When the interval of time is indefinitely small we have, in the notation of the Calculus, dt in place of t, and ds in place of s — S\. The instantaneous speed, or the speed, is then '' = dt ^'^ Sign of Speed. — From (1) we see that if OB or s is numerically greater than OA or s,, the value of v will be positive, or v equals a plus ( + ) quantity. If, however, Si is numerically greater than s, the value of v will equal a negative (— ) quantity. When, then, the distance from the origin, measured along the path, is increasing, the value of v is positive (+). When the distance is decreasing, the value of v is always negative (— ). Moreover, if Si is on the S "4- Si opposite side of the origin from s, we have v = — 7 — . Equation (1) will therefore hold good generally if we take dis- tances from the origin in one direction as (+) and in the other direction as (— ). In such case, if the value of v comes out ( + ) it indicates motion in the assumed (+) direction, and if (— ) it indi- cates motion in the other direction. If t comes out (— ) it denotes time before the start, if (-1-) time after the start or beginning of motion. Speed a Scalar Quantity.— It will be evident from the preceding that the sign of v has no reference to any special direction in space. It simply indicates that the distance along the path from the origin is increasing or decreasing, without reference to the actual direction of the path at any instant. Speed, then, whether uniform or variable, mean or instantane- ous, is independent of direction of the path. A point moving with any given speed has that speed no matter what the shape of the path. Speed, therefore, is a quantity which has magnitude and sign, but is independent of direction. Such a quantity is called a scalar quantity. The student is cautioned here not to confound speed with "velocity," which, as we shall see hereafter (page 43), has direction as well as sign and magnitude. Such a directed quantity is called a vector quantity. CHAP. I.] EXAMPLES— SPEED. 17 Homogeneous Equations.— We have already called attention in the Introduction (page 3j to the fact that the units in all numeric equations are always understood, and when these units are inserted the equation must be homogeneous, that is, every term in it must stand for a quantity of the same kind. When this is not the case some error must have been made in the derivation of the equation, and the relations indicated by it are incorrectly stated. By simply inserting the units, then, in each term of any numeric equation we can at once check the result arrived at and often discover without further investigation if the result is incorrect. Thus suppose that the result of some investigation is expressed by 3s + 2« = lOu. Without reference to the various steps by which this result may have been reached, we can at once say that the result is incorrect. Thus if we insert the vmits, we have Bs[L] +2tlT-\= lOv^^, and we see at once that the quantities in each term are not the same. The equation is not homogeneous. If, however, we had Ss + 2tv= lOvt, this equation is homogeneous, because when we insert the units we have 3s[L] + 2t[T]v^-^- = 10v^^t[T], or Ss[L] + 2tv[L] = 10vt[L]. Here all the terms are quantities of the same kind, and the equation is homogeneous. The relation expressed by it is possible, that ex- pressed by the first is impossible, because we cannot add and sub- tract quantities of different kinds. It does not follow that the relation 3s + 2tv = lOvt is correct. It may still have been incor- rectly deduced. All we can say is it is not on its face absurd, while 3s + 2t = lOy is manifestly so. The student should make it a rule to first test in this manner any equation the truth of which is suspected, as it may often save him the trouble of examining in detail the entire investigation by which it has been deduced. If, however, it stands this test, then the derivation must be examined also. ^ EXAMPLES. ><(1) Water issties from an orifice having an area of cross-section denoted by a, with a speed of v. If the discharge in cubic feet is denoted by q, criticise the foi^mtda q = av. Ans. Since a is the area of cross-section, its unit must be the unit of area, as, for instance, one square foot. The unit of v is the unit of speed, or one foot per second. The unit oi q by statement must be the unit of volume, or one cubic foot. We have then ., V ft. cu. ft. q cu. ft. = a sq. ft. X z = «« :; ^ ^1 sec. 1 sec. The equation is therefore not homogeneous. We have cubic feet on one side, equal to cubic feet per second on the other. If, however, q denotes discharge in cubic feet per second instead of discharge, then the equation becomes homo- geneous. 18 KINEMATICS — GENERAL PRINCIPLES. [CHAP. I« Even then it does not follow that the equation is correct as compared with fact. The actual discharge per second may be less than given hy g = av. But the equation as corrected is homogeneous and may be correct, whereas- before we know it is incorrect, because it states an impossible equality be- tween unlike quantities. (2) A passenger sitting in a railroad car counts n rails passed V over in 20 seconds. If the length of a rail is 29i feet, what is the &peed in miles per hour f Ans. n miles per hour. The number of rails in 20 seconds is the number of miles per hour. (3) If the radius of the earth is 4000 miles, and if it describes a path of 600 millions of miles in 365i dags, find (a) the speed of a point on the equator with reference to a fixed point at the equator, , Y disregarding the motion about the sun ; (b) the speed in the path with /X reference to a fixed point in the path, disregarding the motion of rotation. Ans. J^;^535.89 ft. per sec; (6) 19.01 miles per sec. nearly. U:^If the unit of speed is taken at 30 feet per second, and the unit of length at 20 inches, what should be the corresponding unit of time f Ans. [ F] = Y^ ^^ *^® ^^^* °^ speed mast always equal the unit of length _ [L] _ 20 in. r^.,_ 20in.X l sec._ 1 [ V]~ 30_ft. •'"'l-^ J - 30 X 12 in. ~ 18 ^^'^' 1 sec. (5) If 1 minute is the unit of time adopted, and 1 decimeter per second is the unit of speed, what is the unit of length f Ans. IL] = [V]XIT] = ^-^' X 60 sec. = 60 decuneters. L^€)Tf> The distance of a moving point from a fixed point measured in its path is given by s — at + bf, where s is the number of feet passed over in the number of seconds t. (a) What is the unit of a and b f (b) What is the mean speed between the beginning of the Qth and the end of the 12th second from the start f (c) What is the instantaneous speed f Ans. (a) In order that the equation may be homogeneous, a should be given in ft. per sec. units and b in ft.-per-sec. per sec. units, (b) For ^ = 5 sec. the space passed over is 5a -\- 25b. For f = 12 it is 12a -|- 144&. The distance passed over in the interval 12 — 5 = 7 sec. is 7a -\- 1195. Hence the mean speed is — ^ — a-^ 176. (c) We have for t = ti Si = ati -)- bti^ and hence s — Si = a(i - «,) + Ht^ - ti^) or 7— A" = a-\-b{t-{- U). When the in- t — ti terval of time f — <, is indefinitely small the instantaneous speed is a + 36^. ds By Calculus —= a -|~ 26^ J^^j M >f?) The distance of a moving point from a given point in its path /u^^'^^i-^- is given by s =^ 3 + St, where s is the number of feet passed over in the number of seconds t. (a) What is the unit for 3 and 8 ? (b) What is the mean speed f (c) What is the instantaneous speed f Ans. (a) The number 3 should stand for 3 ft., the number 8 for 8 ft. per sec. (6) When t = 0, the initial distance is Si = 3 ft. Therefore — - — = 8 ft. ■ per sec. (c) Since this is constant, the instantaneous speed is the same. Or CHAP. I.] EXAMPLES — SPEED. 19 S — we can write — = 8 ft. per sec. whicli we see is independent of the interval ^ (8) Suppose the distance is given by s = 7t + 8<^ (a) What is the mean speed and the instantaneous speed ? (6) What is the mean speed between the beginning of the 10th and the end of the 12th second f (c) What is the instantaneous speed at the end of the Gth second f Ans. (a) -^ z=7 -^ 8{t -\- ti) = mean speed. 7 -j- IQt = instantaneous t — ti speed, (b) 175 ft. per sec. (c) 103 ft. per sec. A^ (9) A train runs 40 miles per hour for half an hour, 30 miles an ^ hour for 20 minutes, and 36 miles an hour for 40 minutes. Find its m,ean speed. Ans. 36 miles per hour. (10) A point moves in a circle whose radius is 25 feet and makes 6 revolutions in 15.708 seconds. What is the mean speed f Ans. 60 ft. per sec. ^ tA^ (11) Reduce (a) 60 feet per minute to centimeters per second; (b) 1 Lo^ kilometer per hour to centimeters per second ; (c) 36 feet per second to yards per minute ; {d) 10 yards per second to kilometers per hour. , , 60 ft 60 X 30.479 cm. Ans. (a) r- — ^ — = rr^ = 30.479 cm. per sec. 1 mm. 60 sec. ,^, 1 km. 100000 cm. „„ „ (6) :r-^r— = -^^zr^ = 27.7 cm. per sec. 1 hr. 3600 sec. , , 36 ft. 12 yds. „„„ _, ,^ 10 yds. 10 X 0.0091438 km. „„ „-^„„ , id) —- — = — - — J — r — = 32.9177 km. per hour. 1 sec. aSTTTf '^'"• (12) Compare the magnitudes {a) of the foot per second and the mile per hour ; (b) of the mile per hour and the yard per minute. 1 mile , , Ihr. 5280 ft. 1 sec. 5280 22 „ , ., Ans. (a) -TT-s— = 5^7^7^ X -r-sr = 557^ = T^- Hence 1 mile per hour 1 ft. 3600 sec. 1 ft. 3600 lo 1 sec. 22 is to 1 ft. per sec. as -- to 1, or as 1.466 to 1. 15 1 mile ,,, ThrT 1760 yds. 1 min. 1760 „^, „ . m (6) - — ^ = „-, ■ — X . , = -577- = ^H. Hence 1 mile per hour 1 yd. 60 mm. 1 yd. 60 1 min. is to 1 yd. per min. as 29^ to 1. (13) A point describes 50 feet in 6 minutes and another point de- scribes 50 centimeters in 6 seconds. Compare their mean speeds. 50 ft. 6 min. 50 X 30.47945 cm. 6 sec. 30.47945 ^ ^^„ „ Ans. — = ^^ X ^7^ = — wiT— = 0.508. Hence the 50 cm. 360 sec. 50 cm. 60 6 sec. mean speed in the first case is to that in the second as 0.508 to 1. (14) A man h feet in height ivalks along a level street at a uniform C<x^' fSS^ 20 KINEMATICS — GENBEAL PRINCIPLES. [CHAP. I. ^ speed of V miles per hour, in a straight line from an electric light I ft. tn height. Find the mean speed of the end of his shadow. Ans. I — h : vt :: I : X. Hence I l-h V = speed required. (15) A passenger in a railroad car moving ^l-- with uniform speed counts 50 telegraph poles Jfr at equal intervals of 100 ft. passed in one minute. What is the mean speed of the train f Ans. 56.8 miles per hour. (16) Three planets describe paths which are to each other as 15, 19, and 12, in times which are as 7, 3, and 5. Find their comparative Ans. 225, 665, and 252. (17) Two bodies A and B describe the same path in the same direc- tion, with uniform speeds v and vf , and at the start the distance be- tween them is a. Find the time t when they will be at the distance b in the path, and the distance of each from the initial position of A at the end of that time. Ans. Take, as the example requires, the position of A, when < = 0, as the origin, and let distances in front of this origin be (-|-) and behind it be (— ). Then for the distance of A from the origin at the end of any time t we have « = vt. For the distance of B from the origin at the end of the same time, if B is initially in front of the origin A, we have s' = a-{- v't; if B is initially behind the origin A, s' = — a-\-v't. In general, then, «'= v't ± a, where the (-|-) sign is taken for a, when B is initially in front, and the (— ) sign when B is initially behind the origin A. We have, then, & — s =: vt — v't '^ a. But by the conditions of the problem s — s' =b. Hence b = vt — v't ^ a, or t = b ± a Substituting this value of t, we have b ± a 8 = V ; s' = ± a-\-v' ,b±a v'h ±a/o •where the (-{-) sign or (— ) sign for a is taken according as B starts ahead of or behind A. (18) In the preceding example, suppose the bodies move in the path in opposite directions. Ans. For the distance of A from the origin at the end of any time t, we have, as before, s = vt. If B is initially in front of A and moves in the oppo- site direction, we have s^ = a — v't. If B is initially beJiind A, we liave s' = — a — v't. In both cases, then, s' = — v't ± a, where, as before, the (+) sign is taken for a when B is initially in front of A, and the (— ) sign when B is initially behind A. We have, then, s — s' = b = vt-\-v't T a ; hence t b ± a v-\-v" b ± a v-\-v' ' ±av — v'b V -{- v' (19) Required the time when the two bodies are together. Ans. In this case 6 = 0, and ±a V ± v' v ± v" where the (+) sign or (— ) sign is to be taken for a according as B starts ahead or behind A, and the {-\-) sign or (— ) sign for v' according as the bodies move in opposite or in the same directions in the path. CHAP. I.] EXAMPLES— SPEED. 21 (20) Give finally the general solution. . , b ± a b ± a , ± av ± v'b Ans. t = -. s = v— , 8 = — , V ± V V ± V V ± V ■where a has the (+) or (— ) sign according as B is in front or beliind A, and d' has the (-(-) or (— ) sign according as the bodies move in opposite or in the same directions in the path. (21) Two bodies A and B move in the circumference of a circle of length c, with uniform speeds v and v', the distance apart at the beginning of the time being a. Find the time of the nth meeting, the space described by A and B, and the interval between two successive meetings. Ans. Placing 6 = in the value for t in example 20, we have ± <i ti = = time of the first meeting. v ± V Then t^ = ,— = time of t lie second meetinsf, V ±v' ^ ti = —, — = time of the third meeting. v ± V In general, tn = , = time of the nth meeting. ° V ± V ± a4-(n — l)c , ■. . ■. A Also, s = vtn = •» '— ^-7 — = space described by A, V ±v' ± av' + T(n — 1)C -v J V r> and sT a = ■ — —, = space described by B. V ± V The interval between two successive conjunctions is c 02 — ti ^ fa — fa — 7 • V ± V We take (+) or (— ) sign for a according as B is in front of or behind A at start, and (+) or (— ) sign for v' according as the bodies move in opposite or same directions in the path. (22) When the earth is in that part of its orbit nearest to Jupiter^ an eclipse of one of Jupiter'^s satellites is seen 16 min. 30 sec. sooner than when the earth is most remote from Jupiter. The radius of the. earth's orbit being 92390000 miles, what is the speed of light f Ans^^,>86000 miles per sec. ^f!23) A train of cars moving with a speed of 20 miles an hour had been gone three hours, when a locomotive ivas dispatched in pursuit, with a speed of twenty-five miles an hour. Find the time of meeting, the speeds being maintained uniform during the time. Ans. t = \2 hours. (24) Had the trains in the preceding example started together and moved in opposite directions around the earth, 24840 miles, in what tim£. would they meet f Ans. 23 days. (See example 21.) (25) It is just one o^clock by a clock. Find the time elapsed when the minute and hour hands will be together. Ans. 5y\ minutes. (See example 21.) (26) The daily motion of Mercury in his orbit is 4°. 09239; tJiat of 23 KINEMATICS— GENEKAL PRINCIPLES. [CHAP. I. FewMS 1". 60216; that of the earth 0° .985Q3. What are the intervals between the epochs at which Mercury and Venus respectively will be in the same direction from the sun as the earth ? Ana. 115.876 days and 583.913 days. (See example 31.) (27) A man caught in a shower in which the rain fell vertically, ran ivith a speed of 12 feet per sec. He found that the drops appeared \\ to strike his face at an angle of 10° with the vertical. What was the speed of the drops f Ans. 13 tan 80° = 68 ft, per sec. (28) When the path of the earth in its orbit is perpendicular to a line drawn from a star to the earth, the direction of the star appears to make an angle of 20". 445 with the perpendicular to the path of the earth. The speed of the earth being 68180 miles per hour, ivhat is the speed of light f Ans. 191030 miles per sec. (29) Compare the speeds of two locomotives, one of which travels 897f miles in llf hours and the other 2Q2j% miles in 8f hours. yx Ans. 91 to 81. / / ' cut ^^^ -^ ^^^ makes a circuit of 4i miles in one hour, stopping at V^ 'x/i.o/^*^*' stations Jive minutes and two minutes respectively, and making \x^o^j^ o -((^^Q^^^y QffiQp stoppages of an average duration of 10 sec. each ; find AVyzVC i^^ average speed, ctl^ **^'?*1_ J Ans. Jrr44 miles per hour. uro^'^tj'^'"''^ U-<^) An ordinary train takes ten hours to a certain trip, besides two hours in all of stoppages. The express goes 50 per cent faster, and makes the trip in 4 hours less. What time does it lose in stoppages ? Ans. 1 hour 30 min. (32) A man rides a certain distance and walks back in six hours. He could ride both ways in 3i hours. How long would it take him to ivalk both ways ? Ans. 8^ hours. (33) The speed of the periphery of a wheel 12 feet in diameter is 6 feet per sec; find the revolutions per minute. Y y Anar^.S rev. per min. ^ ^ < v ^ 1^34) A person inquiring the time of day is told that it is between ^^ ^ ' ^ \- 5 and 6 o^clock, and that the hour and minute hands are together. Find the time. '/ ' '^v-*' V/^ Ans. 5 hr. 37 min. 16^*^ sec. (35) Find the number of revolutions per mile made by a wheel 4i feet diameter. Ans. 373 rev. per mile. hi (36) How soon after 8 o'clock are the hour and minute hands ' directly opposite f Ans. 10^^ min. (37) Two men walk opposite ivays round a circular course. ^ They .ji meet for the first time at the north point, the sixth time at tJie east ' point. Where will they meet for the sixteenth time, and what are the relative speeds f Ans. At the w^est point. Ratio of the speeds, 19 to 1, or 3 to 1. 'CHAP. I.] EXAMPLES — SPEED. 23 (38) A courier starts from a given point with a speed of b miles in a hours. After n hours a second courier, travelling at the rate of d miles in c hours, sets out from a point q miles ahead or behind the first point, and travels over the same route. In what time will the second courier overtake the first f (See example 19.) Ans. — jC hours, where the (— ) sign is taken if q is ahead and the (+) (X/Q/ — CO * sign if q is behind. (39) A hollow ball A floating upon the surface of a stream is observed to have a speed of Vo. If A is united by a thin wire to another ball B which sinks in water, the speed of A is now observed to be vi. Find the speed Vi of the water at the depth of B. Ans. We have the combined speed v = -^-P— \ hence »i = 2» — ■Oo- :fo 'At.t Kt^^f--' CHAPTEE II. RATE OF CHANGE OF SPEED. EQUATIONS OP MOTION UNDER DIFFERENT RATES OF CHANGE OF SPEED GRAPHIC REPRESENTATION OF RATE OF CHANGE OF SPEED. Change of Speed.— When the speed of a point is variable, th© difference between the final and initial instantaneous speeds for any interval of time is the total or integral change of speed for that time. Mean Rate of Change of Speed.— The integral change of speed per unit of time is the mean rate of change of speed. Therefore the number of units in the integral change of speed for any interval of time, divided by the number of units in that time, gives the number of units in the mean rate of change of speed. When the mean rate of change of speed varies with the interval of time, it is variable. When it has the same magnitude, no matter what the interval of time, it is uniform. Instantaneous Rate of Change of Speed. — The limiting value of the mean rate of change of speed, when the interval of time is in- definitely small, is the instantaneous rate of change of speed. Rate of change of speed should always be understood as mean- ing the instantaneous rate of change unless otherwise specified. Rate of change of speed may be zero, uniform or variable. When it is zero, the speed is uniform and is the same as the mean speed for any interval of time. When it is uniform, the rate of change of speed is the same as the mean rate of change for any interval of time. When it is variable, the mean rate of change has different values for equal intervals of time. Dimensions of Unit of Rate of Change of Speed.— Let us denote the rate of change of speed by A and its unit by \A J and the number of units by a, so that A = a[A\ Then if v^\V] and u[F] are the initial and final instantaneous speeds and t[T] the corresponding interval of time, we have the integral change of speed {v — Ui)[F], and by definition the mean rate of change of speed is a[A]- ^j . We shall always have the numeric equation V — Vi t if we take [A] = [^ = [^i(pagel6). CHAP. II.] RATE OF CHANGE OF SPEED. 25 The unit of rate of change of speed is then one unit of speed per unit of time, as for instance one foot-per-sec. per sec. Numeric Equations of Rate of Change of Speed.— We have then the numeric equation for the mean rate of change of speed a = —^, (1) where v and th are the number of units in the final and initial instantaneous speeds during the interval of time t units. For the instantaneous rate of change of speed we have the limiting value in the notation of the Calculus dv d^s '' = dt=dt^ ^2> Sign of Rate of Change of Speed.— From (1) we see that if v is numerically greater than Vi the value of a will be positive ( + ), and if tJi is numerically greater than v the motion is retarded and the value of a is negative (— ). The value of a then is positive (+) when the speed increases, and negative (— ) when the speed decreases during the interval of time t. Rate of Change of Speed a Scalar Quantity. — It is evident that the sign of a has no reference to any special direction, but simply shows whether the speed is increasing or decreasing numerically. Rate of change of speed is therefore, like speed (page 16), a scalar quantity. The student is cautioned here not to confound rate of change of speed with "acceleration," which, as we shall see hereafter (page 49), has direction as well as magnitude and sign, and is therefore a vector quantity. WtlT^ EXAMPLES. [) A paint moves at a given instant with a speed of 12 ft. per sec., and at the end of 3 sec. after, with a speed of 16 ft. per sec. What is the change of speed, and the mean rate of change of speed f Ans. +4 ft. per sec. ; -]- 1 ft.-per-sec. per sec. The (+) sign indicates that the speed increases. (2) A train starts from, rest and in 8 m,in. attains a speed of 30 miles an hour. What is the uniform rate of change of speed f Ans. -f" 225 miles-per-hour per hour. (3) Compare the foot-per-sec. per sec. and the yard-per-min. per min. Ans, 1 ft.-per-sec. per sec. is equivalent to 1300 yards-per-min. per min. v(4) Reduce 400 ft.-per-min. per min. to kilometers-per-hour per hour, to feet-per-sec. per sec, and to centimeter s-per-sec. per sec. Ans. 438.9 kil.-per-hour per hour ; J f t. -per-sec. per sec. ; 3.38 cen. -per sec. per sec. (5) Reduce 1 mile-per-min. per min. t9 centim^eters-per-sec per sec. Ans. 44.7. L^:(ff^~lve take 32.2 ft.-per-sec. per sec. as the unit of rate of change of speed, and 5 yards as the unit of length, what should he the unit of time f Ans. [T] = l/g] = 1/3^ 1 sec* = 0.68 sec. ^ 36> KINEMATICS — GENERAL PRINCIPLES. [CHAP. II. * (7) In 15 sec. the speed of a point changes from 400 to 100 ft. per sec. Find the uniform mean rate of change of speed. Ans. — 20 ft.-per-sec. per sec. The minus sign shows that the speed decreases. (8) The distance measured in the path of a moving point from a fixed point is given by s = at + bt'\ where s is the number of feet passed over in the number of seconds t. What is the mean speed, instantaneous speed, mean rate of change of speed, instantaneous rate of cliange of speed f Ans. Mean speed, = a -\-b{t -\- ti). Instantaneous speed, —= a -|- t — fci (it 2bt. Mean rate of change of speed, = 2b. Instantaneous, the same. t — Ti y (9) If the distance is given by s — 2t + Sf + 4t\ what is the mean speed between the origin and final position at the end of 5 sec. f What is the instantaneous speed at the origin and at the end of 5 sec. f What is the mean rate of change of speed between the origin and at the end of 5 sec. f What is the instantaneous rate of change of speed at the origin and at the end of 5 sec. f Ans. Mean speed, *^' = 3 + 3(« + ti) + Mt^ + tt, + ^i'^. If we let «, = when t^ = 0, the time t counts from the origin, and we have for t = 5 mean speed = 117 ft. per sec. ds Instantaneous speed, v = ~ =^ 2^&t-\- 12i*. For t = this gives 2 ft. per sec. For t = 5, 332 ft. per sec. 1) — 7} Mean rate of change of speed, -' = 6 -}- 12{t -f ti). For ^, = and t = 5, this gives 66 ft.-per-sec per sec. Inst9,ntaneous rate of change of speed, —- = Q-\- 2it. For ^ = this gives 6 ft.-per-sec. per sec. For t = 5, 126 ft.-per-sec. per sec. (10) If a point traverses in t units of time a distance measured in the path of s units of length, given by s = -t+ bt', where a and b are constants, what is the mean speed, the instantaneous speed, the mean rate of change of speed, the instantaneous rate of change of speed ? Ans. Mean speed, -' = — — -\-h{t -\- ti). Instantaneous speed, v = — — — --{- 2bt. Mean rate of change, j- = — — bj— ^ + 2o. Instantaneous rate of change, -— = -— -f 26. at f (11) A steamer leaves Liverpool for New York, and a vessel leaves New YorJc for Liverpool at the same time ; they meet, and when the steamer reaches New York, the vessel has as far to go as the steamer had when they met. If the distance is 3000 miles, how far out from Zdverpool did they meet f Ans. 1854 miles. ~ ' '• OHAP. II.] KATE OF CHANGE OF SPEED. 27 > (12) A tvalks at the speed of 3| miles per hour and starts 18 minutes before B. At what speed must B walk to overtake A at the ninth milestone f Ans. 4.29 miles per hour. (13) A tourist left behind by his companions, wishes to rejoin them on the following day. He knows they are 5 miles ahead, ivill start in the morning at 8 o^ clock, and will walk at the rate of 3i miles an hour. When must he start in order to overtake them at 1 o^clock p.m., walking at the rate of 4 miles an hour, and resting once for half an hour on the road ? Ans. 7 h. 12 m. a.m. (14) A man walking 4 miles an hour meets 20 street cars in an hour and is overtaken by 4. What is the average speed of the cars, and ivhat is the average distance between successive cars f Ans. 6 miles per hour ; ^ mile. (15) J. starts from a raihvay station, loalking 5 m,iles an hour ; at the end of an hour B starts, ivalking 4 Tniles an hour. At the end of another hour a train starts and passes A 25 minutes after it passes B. Find the speed of the train. Ans. 20 miles an liour. ^ffo) A passenger-train going 41 rniles an hour and 431 feet long overtakes a freight on a parallel line. The freight-train is 713 feet long and is going 28 miles an hour. Hoio long does it take the pas- senger-train to pass ? Ans. 1 minute. (17) In a mile race A gives B 50 yards. B passes the line 5 TThinutes after the start. A passes it 5 seconds later. Which would win in an even race, and by what distance f Ans. A by 21^ yards. Equations of Motion of a Point under DiflFerent Rates of Change of Speed. — The rate of change of speed may be zero, uniform or variable. When variable, it may vary according to any law. (a) Rate of Change of Speed Zero. — When the rate of change of speed is zero, the speed is uniform, and the instantaneous speed at any instant is equal to the mean speed for any interval of time. In this case, if Si is the number of units m the initial distance, measured along the path from the origin, and s the number of units in the final distance, we have s — Si v= — 7 — , or vt = s—8i (1) This equation will be general if we take distances from the origin in one direction as ( + ) and in the other direction as (— ). In such case, if the value of v comes out ( +) it denotes motion in the assumed ( +) direction; if (— ), it denotes motion in the other direction. If t comes out ( + ) it denotes time after, if (— ) time before the start, or beginning of motion. (6) Rate of Change of Speed Uniform. — When the rate of change of speed is uniform, the instantaneous rate of change of speed at any instant is equal to the mean rate of change of speed for any interval of time. In this case, if v and Vi are the number of units in the initial and / 28 KINEMATICS — GENERAL PRINCIPLES. [CHAP. II.. final instantaneous speeds for any interval of time t units, we have for the uniform rate of change of speed a = — ^ — (2) The value of a is (+) when the speed increases and (— ) when it de- creases during the interval of time t. From equation (2) we have V = Vi + at (3> The average speed during the interval t is V+Vi 1 mean speed = — ^ — = vi + ^at (4> The distance (s — Si) betiveen the initial and final positions, de- scribed in the time t, is the mean speed multiplied by the time, or s — s, = — 2 — t = v,t + -^aP (5> Inserting the value of t from (2) we have v" — v," '-'' = ^r- («> Hence v' = Vi' + 2a(s — Si) (7) In applying these formulas, a is positive (+), when the speed in- creases, and negative (— ), when the speed decreases, without regard to direction of motion. If distances s, Si , from the origin, in one direction are taken as positive (+), distances in the opposite direction are negative (— ). Speeds v, v^ are positive ( + ) when motion is in the assumed fositive direction, and negative (— ) when in the other direction, f ^ is minus (— ), it denotes time before the beginning of motion; if plus ( -I- ), time after. By means of these equations, if we have given the initial position of a point moving in any path, its initial speed and uniform change of speed, we can determine its final position and speed and the dis- tance described in any given interval of time. [(c) Kate of Change of Speed Variable.] — If the rate of change of speed is variable, we have from (1), in Calculus notation, for the instantaneous speed, ^ = d?= <^> and from (2), dv <Ps .„, ''=di=dl^'' ^^> and from (8), s - s, = / vdt (10) The preceding equations for constant rate of change of speed can be directly deduced from these three general equations. Thus if we suppose a constant, we have, by integrating (9), v = at -^ C, where C is the constant of integration. When ^ = 0, we have v = v^, and hence C = ^.-t CHAP. II,] RATE OF CHANGE OF SPEED. 29 d N «, . Hence v = r^ -\- at, whicli is equation (3). Inserting this value of v in (8) and integrating, we have s — Vit -}- iat'^ -\- U. But when < = we have « = s, , hence C = «,, and, therefore, s — «, = Vit -{■ iaP. This is equation (5). In any case, if the law of variation of a is given, we can find the relation between v, s and t. Graphic Representation of Rate of Change of Speed. — If we rep- resent intervals of time by distances laid off horizontally along the axis of X, and the corresponding speeds by ordinates parallel to the axis of y, we shall have in general a curve for which the change of y with X will show the law of change of speed with the time. (a) Rate of Change of Speed Zero. — Lay off from A along AB equal distances, so that the distances from ^ to 1, 1 to 2, 2 to 3, etc., are all equal and represent each one second of time, and let AB represent the entii-e time f. y Then at A, 1, 2, 3, and B erect the perpendiculars AM, lb, 2c, 3d, BN, and let the length of each represent the speed at the corresponding instant. Since there is no change of speed, these perpendiculars will all be of equal length, we shall " '''■ •*"• '^■ have AM =lb = 2c^3d = BN ^ t- " = V, and the speed at any interval of time will be given by the ordinate at that instant to the line MN parallel to AB. The space described in any time is given by s—Si = vt. This is evidently given by, the area, AMNB in the diagram. Therefore, the area corresponding to any time gives the space described in that time. (6) Rate of Change of Speed Constant. — Lay off as before the time along AB, and at A, 1, 2, 3, B, the corresponding speeds, so that AM is the initial speed r, and BN the final speed v. Draw MC, be', cd', parallel to AB. Then bb' will be the change of speed in the first sec, cc' the change of speed in the next sec, x and so on. Since these are to be constant, NM is a straight line, the ordinate to which at cc dd' sec 1 sec. any instant will give the speed at that instant. The rate of change of speed is then = - But = — ^' = a. Hence the rate of change of speed is the tan- gent of the angle NMC which the line MN makes with the horizontal. Hence a = — or NC = at. The distance described in the time t is from equation (5) given V -i- "V by — K-^f- But this is the area of AMNB. Therefore, the area corresponding to any time gives the space described in that time. We have then directly from the figure, since NC = at, (ft' ^-^t = vd + \nC xt = va + lat\ 30 KINEMATICS — GENERAL PRINCIPLES. CHAP. II, If V is greater than v, a will be negative, and the line MN is in- clined Delow the horizontal MC. [(c) Kate of Change of Speed Variable.]— If the rate of change of speed is not constant, we shall have in general a curve MNn. The tangent to this curve at any point N makes an angle with the axis of X, whose tangent is -7- = a, equation (9), or the rate of change of speed. The elementary area BNiib = vdt = ds, equation (8), and the total area AMNB = £ vdt = s— »,, equation (10). t-o dv d's When -— = 0, or -— - = 0, dt ' dt^ ^ or a = 0, B b the tangent to the curve is horizontal at the corresponding point, and we have the speed at that point a maximum or minimum, according as the curve is con- cave or convex to the axis of X. EXAMPLES. [) The speed of a point changes from 50 to 30 ft. per sec. in pass- ing over 80 ft. Find the constant rate of change of speed and the time of motion. e*_B,, 900-2500 ._ ,^ rpi, • = — " — — — =. — 10 ft.-per-sec. per sec. The minus Ans, a = 2(« - «j) 2 X 80 sign indicates decreasing speed, t — v^ _ 30 - 50 ~ ~ -10 = 2 sec. (2) Draw a figure representing the motion in the preceding exam- ple, and deduce the results directly from it. Ans. Average speed = x — = 40 ft. per sec. Hence ^-30 40« = 80, or t sec. per sec. 2 sec. Also a = 30 - 50 2 = - 10 ft. -per- (3) A point starts from rest and moves with a constant rate of change of speed. Shoio that this rate is numerically equal to twice the number of units of distance described in the first second. Ans. We have t 2(8 -Si) 1 and ®, =0; hence from eq. (5) -Si 1 1 sec 1 sec. , which is numericaUy equal to 2(s — s,). ^J^In In an air-brake trial, a train running at 40 miles an hour was stopped in 625.6 ft. If the rate of change of speed was constant during stoppage, what was it f 40 y 5*^80 Ans. From eq. (6), we have for » = 0, s — s, = 625.6, and », ^1" _ (40 X 5280)" 2 X 625.6 (60 X 60)" X 2 X625.6 The (— ) sign shows retardation. 60 X 60 ' = — 2.75 ft -per-sec. per sec U»<1 A point starts with a speed r, and has a constant rate of change of speed — a. When will it come to rest, and what distance does it describe f CHAP. II.] RATE OF CHANGE OF SPEED. 31 Ans. From eq. (3), when « = 0, we have v^ — at = 0, or t ~ — . Prom eq» <"»-'■ = 5 '=-5^- \*^ A point describes 150 ft. in the first t/iree seconds of its motion and 50 ft. in the next two seconds. If the rate of change of speed is constant, when will it come to rest f When will it have a speed of 30 ft. per sec. f g Ans. From eq. (5) we have for « — «j = 150 and t = S, 150 = 3®, + o^ ; and 26 for 5 — «, = 300 and t = 5, 200 = 5®, -|- -^a. Combine these two equations and we have a = — 10 ft.-per-sec. per sec, and v^ = 65 ft. per sec. From eq. (3), if B = 0, we have 65 — 10^ = 0, or ^ = 6.5 sec. From eq. (3) we also have iiv = 30, 30 =r 65 - lOt, oTt = 3 5 sec. v-<?r^ A point whose speed is initially 30 meters per sec. and is decreasing at the rate of 40 centimeters-per-sec. per sec., moves in its path until its speed is 240 meters per minute. Find the distance traversed and the time. Ans. We have Vi = 30 and ® = 4 meters per sec, and a = — 0.4 meters- per-sec per sec. From eq. (6) s — Si = 7rQ~ — ^^^^ meters. From (3) we — O.o have 4 = 30 - 0.4t, or t — 65 sec. LJi&fA point has an initial speed of Vi and a variable rate of change of speed given by + kt, where k is a cotistant. What is the speed and distance described at the end of a time t f dv kfi Ans. From eq. (9) we have a = -— = kt, and integrating, v = — - -j- C. If, (it i kt'* when t =-0, we have b = «, , we obtain G = Vi , and hence v = v^ ~\ — —. kPdt kt^ From eq. (8) ds = vdt = v.dt H k-. Integrating, s = ® /_)_-_ -j- C. If, a 6 kt^ when ^ = 0, we have s = 0, we obtain C = 0, and hence s — v,t-{- -^-. 6 (9) A point has an initial speed of 60 ft. per sec. and a rate of change of speed of + 40 ft.-per-sec. per sec. Find the speed after 8 sec. ; the time required to traverse 300 ft.; the change of speed in traversing that distance ; the final speed. Ans. From eq. (3) we have ® = 60 + 40 X 8 = 380 ft. per sec. From eq, (5) we have 300 = 60f -\- 20<^ or t =±y— -- = + 2.65 sec or - 5.65 sec. The first value only applies. _ From eq. (3) we have v - v, - 40t - 20(± V69 - 3) = -f 106 ft. per sec. or — 226 ft. per sec. The first value onlv applies. We have for the final speed « = + 166 ft. per sec. or — 166 ft. per sec. The first value only applies. That is, the point starts from A with the speed v^ = 60 ft. per sec. and de- scribes the path AB = 300 ft in ^ = 2.65 sec, the speed at B being v = 166 ft. per sec. ~ *" In order to interpret the negative values ^ 300/?7 obtained, we observe that v =z — 166 ft. per 'A v^-^ + eo sec. means that the point moves in the opposite t - 2.65 direction. KINEMATICS— GENERAL PRINCIPLES. [CHAP. II. f-5.65 Let tlie point then start from B in the opposite direction with the speed «. = — 166 ft. per sec. Then from eq. (3) we have « = — 166 -|- 40^. We see that when t = 4.15 sec, v -- 0, and thejjoint has passed to some point P, where the speed is zero. This point is the turning- point. For t greater than 4.15 sec. v be- comes positive; that is, the point moves back towards B, and arrives at a point A where the speed is ® = -f- 60 in the time given by 60 = 40^, or t — 1.5 sec. The entire time from B to P and back to J. is ^ = 5.65 sec. This is the time given by the negative value of t in the example ; that is, it is the time before the start, during which the point moves from B to P and back to A. The change of speed ® — Wi is -f- 60 -f- 166 = + 226, which is the negative value in the example. For the space BA described between the initial and final positions, we have '^^^^t or + ^Q ~ ^ jg -^ 5 65 ^ _ 300 ft., the (-) sign showing that the dis- tance is on the other side of the origin, from the case of the example. We see then that our equations are general if we have regard to the signs of v, Vj ,«,«,, and a. (10) If the motion in example 9 is retarded, find (a) the distance described from the starting to the turning point ; (6) the distance de- scribed from the starting-point after 10 sec, the speed acquired and the distance between the final and initial positions ; (c) the distance described during the time in which the speed changes to —90 ft. per sec, and this time; (d) the time required by the moving point to return to the starting-point. Ans. The initial speed is v^ = -f-60 ft. per sec, the rate of change of speed is a = — 40 ft.-per-sec. per sec. Let the time count from the start at A, so that Si = 0, when ^ = 0, and let dis- tances and motion from A towards P 9 ~^^:^ ^.^ be positive. (a) We have from eq. (6) for the distance from A to the turning-point P, where v — 0, . -«,» -3600 , ._^ - = -j- 45 ft. p 17 — 2a -80 (b) From eq. (5) we have for the distance between the initial and final posi- tions after 10 sec. AB = s = v^t-{--at^ = 60 X 10 — 20 X 100 = - 1400 ft. The minus sign shows distance on left of A. The total distance described is then 1490 ft. The speed acquired is given by eq. (5) — 1400 = ^-^ — X 10 OT V = — 340 ft.-per-sec. The minus sign shows motion from A towards B. (c) From eq. (6) the distance between the initial and final positions, when '_ . _ ^'' - •"■' _ 8100 - 3600 2a ~ -80 ft. The minus sign shows that C is on left of A. The total distance described from the start is then 56.25 + 90 = 146.25 ft. The time, from eq. (5), is _ 56.25 = ~^^ + ^\ or t = 3.75 sec. (d) The time to reach the turning-point, as we have seen, is 1.5 sec. The time to return is, from eq. (5), 45 = — ^ = 1.5 sec. The entire time to go and return is then 8 sec. the speed is — 90 ft. per sec, is ^C= « = = - 56.25 ■CHAP. II.] RATE OF CHANGE OF SPEED. 33 ^ (11) A railway train runs at a speed of 20 miles an hour, and its speed is increasing uniformly at the rate of 14 feet-per-min. per min. Pind its speed after li hours, and the distance traversed in that time. Ans. 14 feet-per-min. per min. =; 9.54 miles-per-bour per liour. From eq. (3) 17 — 20 -|- 9. 54 X 1.5 = 34.3 miles per liour. From eq. (6) the distance de- ., ^ . 34 3» - 20« ,^ „ ., scribed is —r- t-^-t- = 40.7 miles. 1^^^^ 2 X 9.54 (12) A railway train moving with a speed of 50 miles an hour has the brakes put on, and the speed diminishes uniformly /w 1 minute, when it is found to he 20 miles per hour. Find the rate of change of speed, and the distance traversed. Also the time in which it ivill come to rest and the distance traversed. Ans. From eq. (3), a = = — — — 1800 miles-per-hour per hour. 60 From eq. (6) the distance traversed is 5^-- — = j^ mile. In order to come — obOO 1^ to rest, t = = — rKP{h = oS liour = 100 sec. The distance in coming to (I — loOO 00 -V* -2500 25 ., (13) Express a rate of change of speed of 500 centimeters-per- ^econd per second, in terms of the kilometer and minute. Ans. 18 km.-per-min. per min. C/^ (14) The speed and rate of change of speed of a moving point at a certain moment are both measured by 10, the foot and second being the units. Find the number measuring them when the yard ana minute are the units. Ans. 200 yards per min. ; 12000 yards-per-min. per min. f^ (15) What is meant when it is said that the rate of change of speed of a point is + 10, the units being foot and second f If the point were moxnng at any instant at the rate of 7i feet per second, after what time would its speed he quadrupled f and what distance would it de- scribe in that time f Ans. 2i-sec.; 42.1875 ft. (16) A body describes distances of 120 yards, 228 yards, 336 yards, in successive tenths of a second. Show that this is consistent with constant rate of change of speed, and find the numerical value if the units are a minute and a yard. Ans. 38880000 yards-per-min. per min. ^ (17) Anoint is moving at the rate of h feet per sec, a quarter of a ' minute after at the rate of 5Q feet per sec., half a minute after at 95 Jeet per sec. Shoio that this is consistent with a constant rate of change of speed, and find its value. Ans. 3 ft.-per-sec. per sec. CHAPTEE III. DISPLACEMENT. RESOLUTION AND COMPOSITION OF DISPLACEMENTS. Displacement. — The total change of position, measured in units of length, of a point, between its initial and final positions, without reference to the path described, is the linear displacement of the point. Thus if a point moves from the position Ai to the position Aa, the distance AiAi is the linear displacement, no matter what the path may have been from Ai to ^2. The term displacement always means linear displacement unless otherwise specified. Line Representative of Displacement. — The displacement of a {)oint is therefore completely represented by a straight line. The ength of the line gives the magnitude of the displacement, and the direction as denoted by an arrow gives the direction of the dis- placement. Thus the straight line AiAi represents by its length the magnitude of a displacement, and the arrow shows that the displacement is in the direction from Ai to A2. A displacement then has both magnitude and direction, and such a quantity is called a vector quantity. All vector quantities can Ije represented thus by a straight line.* Relative Displacement. — Since the displacement of a point is change of position, it can only be determined by reference to some chosen point of reference. Thus if Ai and Aa are the initial and final positions of a moving point A, and Bi is some chosen point of refer- Bjf; Ts/ ence, we call AiA, the displacement of A with \ /^y reference to Bi , because to an observer at B, \ / \^ the point A would move from ^1 to A-i in the \ / \. direction A,Ai. B Aj But to an observer on the moving point A , the point B, would appear to move from Bi to Bi , and this is really the displacement of B relative to A. That is, any change in the relative position of two points A and * As we shall see liereafter, linear velocity and. acceleration, angular ve- locity and acceleration, moment of linear velocity and acceleration, moment of angular velocity and acceleration, are all vector quantities, and the same prin- ciples apply to all of them as to displacements, 84 CHAP. III.] DISPLACEMENTS. , 35 B may be regarded either as a displacement of A relative to Boras an equal and opposite displacement of B relative to A. Relative Displacement— Two Points.— If then the straight line AB represents the displacement of a point A with reference to a point JS, the equal and op- ^ posite line BA represents the displacement of B relative to A. We shall always denote therefore relative displacement by a line, the length of which gives the magnitude of the displacement, the arrow its direction, and the letters at the end the two points. Thus in the figure one line gives the displacement of A relative to JB, the other gives the displacement of B relative to A. Relative Displacement — Three Points —Triangle of Displace- ments.*— Let a moving point A have the dis- ^"s;— ^ ^ placement AB with reference to a point B, / J><'' \ " ^^^ ^^ *^® same time let B have the displace- / r""'''^^^'^ ment BC with reference to a point C, , and let qZ. >5.i ~>c B^ , Bi be the initial and final positions of the \ ^'^' point B. v^ Then it is evident that since the point has * the displacement AB with reference to B, and at the same time B moves from B^ to Bi , the point moves from A to C, and AC is the displacement of A with reference to C. Conversely, from the preceding article, CA is the displacement of C relatively to A. Hence, if two sides of a triangle ABC taken the same way round represent the displacements of A relative to B and B relative to C respectively, the third side taken the opposite ivay round will repre- sent the displacement of A relative to C, and taken the same y:>aUr , J round, the displacement of C relative to A. Ar^^ ^s/r^ i) r This is called the principle of the triangle of displacement^, xfc--' ^ ^ evidently makes no difference whether the displacements are simultaneous or successive. The same principle holds true in. both cases. Polygon of Displacements. — Let AB, BC and CD be the given displacement of A relative to B, B relative to C, and C ^3 relative to D. Then if we lay off succes- siyely AB and BC in given direction and magnitude, the line AC gives the displace- ment of A relative to C. If we lay off CD, the line AD gives the displacement of A relative to D. - Hence, if any number of displacements in the same plane are represented by the sides of a plane polygon, ABCD, etc., taken d the same way round, the line AD which closes the polygon taken the opposite way round will give the magnitude and direction of the re- sultant displacement of the first point relative to the last, and taken the same way round, of the last point relative to the first. This principle is called the polygon of displacements, and it evi- * We shall see hereafter that all the principles which follow in this chap- ter relating to displacements hold equally good for linear and angular veloci- ties and accelerations, for the moments of linear and angular velocities and ac- celerations, and for forces and moments of forces. B C \/ 36 KINEMATICS — GENERAL PRINCIPLES. [CHAP. III. dently holds good whether the displacements are successive or simultaneous. Resolution of Displacements. — By the application of these prin- ciples a given displacement or any number of given displacements may be resolved into two components in any two given directions. Thus suppose the displacement of A relative to -B to be given by the line AB. We can re- solve this displacement into components in any two directions given by the arrows a and 6, by drawing lines from A and B parallel to these directions till they intersect at some point C. Then the displacements AC and CB taken the other way round are the component displacements of AB in the required directions. If any number of displacements AB, BC, CD, etc., ^ are given, we have the resultant displacement AD, and this displacement can be resolved as before into any two directions required. Rectangular Components.— When a displacement is thus resolved into two directions at right angles, the components are called rectangular components. Unless otherwise specified, when we speak of the components of any displacement, the rectangular components are always under- stood. The Component of the Resultant is equal to the Algebraic Sum of the Components of the Displacements. — It is evident that the resultant of any two given displacements is equal _^B to the algebraic sum of their components in the direction of the resultant. For if AC and CBaxe the given displacements, the resultant AB is equal to the sum of the com- ponents AD and DB. So also for any number of displacements, the resultant AE is equal to the algebraic sum of the components of the displacements in the direction otAE. The component in any given direction, of this resultant itself, is then equal to the algebraic sum of the components of the displacements in the same direction. Thus the projection of the resultant AE upon the line OP, or the component of AE in the direc- tion OP, is the algebraic sum of the components Y of AB, BC, CD and BE in the direction OP. Components not in the Same Plane. — The same principles apply for components not in the same plane. Thus if OA is a given displacement, and we draw AB perpendicular to the plane of XZ meeting it at B, we have the components OB and BA. Again, we can resolve OB into the compo- nents OC and CB. The components then are OC, CB and BA. Sign of Components of Displacement. — A sign of (4-) or (— ) pre- fixed to the magnitude of a component displacement indicates direction. Thus for three rectangular axes OX, OY, OZ, a com- CHAP. III.] DISPLACEMENTS. 37 ponent displacement in the directions OX, OY, OZ is positive, and in the opposite directions negative. If polar co-ordinates are used, the com- ponent displacement along the radius vector is positive when away from the pole, nega- tive when towards the pole. Evidently, then, we measure angles in the plane XY from OX around towards 0Y\ in the plane YZ from OY around towards OZ; in the plane ZX from OZ around towards OX, as shown by the arrows in the figure. Analytical Determination of the Resultant for Concurring Dis- placements, — When the line representatives meet in a point they are called concurring. All displacements of a single point must be concurring. The magnitude and direction of any number of such concurring component displacements being given, to find expres- sions for the magnitude and direction of the resultant. (a) When there are Two Given Components. — Let OB — di and BC = di be two component displacements making the angle a and in the indicated directions. Then the resultant is OC = dr and is given at once in magnitude from the tri- angle OBC, dr' = {di + di cos ay + (di sin ay — di" + 2d,di cos a + d-y. . (1) The resultant dr makes with d, an angle a, given by cos a = di + di cos aa dr (2) (6) When there are Any Number of Components in any Given Direction. — Take three rectangular axes, OX, OY, OZ, through the point O, and let the component displacements di, di, da, etc., make the angles «r,, /S,, y -^ a^, /S^, r-!, etc., with the axes of X, Y, Z respect- ively. Then we have for the sum of the components in the direction of X, Y, and Z, dx = di cos a-i + di cos a, + . . . = .2d cos a; dy = di cos A + di cos A -f- . . . = 2dcos/3; dz = di cos^i + di cos Vi + . . .= ^dcosy. ) In these summations we must take components with their proper signs as directed in the preceding Article. We have then for the magnitude of the resultant dr, naA-^ dr = Vdx' + dy' + dzK . . . J^-P-r. . (4) This resultant makes with the axes angles a, 6, c, respectively, given by "'(5) d^ r dii dz cos a = 3- , cos 6 = -="-, cos c = --. dr Or <*r ^i "38 KINEMATICS — GENERAL PRINCIPLES. [CHAP. III. The projections on the planes XY, YZ, ZX make angles with X, Y, Z, respectively, whose tangents are Uy dz^ dx ,„, dx CLy CLz Cor. 1. If all the displacements are in one plane, make dz = 0. Then ^_^_ dr = \/dx' + dy' (7) cos a = ~, cos b = ~\ (8) dr dr and, since cos b = sin a, tana = ^ (9) dx Cor. 2. When there are hut two displacements, di and d, take OX corresponding with d, . Then dx = di + d cos a,dy — d sin or, and we have at once equations (1) and (2). ^^ Cor. 3. If the two displacements are in the same direction or in j/V "^ opposite directions, a = or 180°, and dr ^=di ± d. That is, the resultant is the algebraic stun of the displacements. I ,^/-^CoR. 4. If the two displacements are at right angles, « = 90° and ji/sA^"' d d d dr = Vdi' + d', sin a = ^— , sin &=-_—= cos a, tan a = -r- dr dr di Cor. 5. If the two displacements are equal, di = d and dr' = 2d*(l + cos a) = 4d' cos' ^, ./: hence dr = 2d cos ^. . , . d sin a . a . a I Also, sin a = = sm — , hence a = — ri^ /- « , « 2 2 j^" ( 2dcos- CoR. 6. If the two displacements are equal and a = 120", then dr = d, and a = 60°. EXAMPLES. (1) A wheel whose radius is r rolls on a horizontal plane until it turns through a quarter revolution. Find the displacement of the point of the wheel initially in contact with the plane relative to the point diametric- ally opposite. Alls. Let C represent the initial point of con- tact of tlie plane, A the point of tlie wheel in con- tact with C in the plane, B the point of the wheel diametrically opposite A. The displacement of B with reference to C is CHAP. III.] DISPLACEMENTS — EXAMPLES. 39 given by BiBi. y^'A The displacement of A with reference to Cis given by AxA^, and therefore the displacement of C with reference to A is given by A^Ai. We have then, by laying oS these displacements, the displacement ot A relative to B equal t( ? 2£j /^ , making an angle of 45° with the verticUh;^!^.' *tf|L-rfncZ the displacement of the end of the frninute-hand with reference to the end of the hour-hand of a clock, between 3 and 3.30 o'clock, the length of each hand being r. Ans. The displacement of the minute-hand with reference to XII is MiMi , and of the hour-hand with reference to XII, HiH, . Therefore the displacement of XII with refer- ence to ZT is HiHi . Laying off these displace- ments, we have XliSi MR= r -/e - 3 cos 15° — 4 sin 15°, and for the angle of MH with the vertical 2 sin2 7^° sm^JfXII J V'6 — 2 cos 15° — 4 sin 15° (^ (3) A river flows in a direction N. 33° E., and a boat headed directly axiross at right angles to the current reaches a point on the other shore from which the starting-point is found to bear S. 3° W. If the distance from the starting-point is 500 ft., how far has the boat been carried by the current, andivhat is the distance across the river if the banks are parallel f Ans. The course of the boat makes an angle of 30° with the bank. The distance across is 250 ft. ; the distance parallel to the bank 433 ft. I ^^tA^ ■ (4) A point A moves 30 ft. in a given direction relative to a fixed Jj point O. Another point B moves relative to O 40 ft. in a direction Jr at right angles to A's motion. Find the displacement of A relative to B. Ans. 50 ft. in a direction inclined to A'& motion by an angle whose sin JWhTwo points move in the circumference of two circles of radius r arid 2r respectively. Both start from the point of contact. One moves through an angle of n radiants, the other through an angle of Tt * — radians. Find the displacement of either relative to the other. Ans. If one point moves through the angle Tt in the small circle and the other through — in the large, the displacement is 2r 4/5 , making an angle with the vertical whose tangent is 2. If one point moves through the angle it in the large circle and the other through -^ in the small, the displacement is rv'26 and it makes an angle with the vertical whose tangent is 5. IJ^-Jn the preceding example let the radius of the small circle be Ti ahd of the large circle r,. Find the displacement as before. 40 KINEMATICS — GENERAL PRINCIPLES. [CHAP. IIL. ^ if Ans. 'l/4ri'' + 4r2n + 3ri» and tangent 2u + r. Ti V4ra' + 4»"a»*i 4- Sti" and tangent 2ra + ri ri (7) TTie displacement of A relative to B is a distance a towards the south, and relative to C, c towards the west. If C is initially a distance b south of B, find the final position of C relative to B. Ans. We have CB for the displacement of C relative to B. Therefore if Ci is the initial po- sition, C-, will be the final position, CjCo being equal to CB. Hence the distance of C's final position from B is Bd = |/(6 -f ay^^\ and the direction from B to Cj is east of south an angle whose tangent is 6 + a" (8) A's displacement relative to B is a to the west. C^s displace- ment relative to B is c in a direction 30° west of south. Find the displacement of A relative to C. Ans. Displacement is |/a' — ac-{- c" . 2a — c Tangent of the angle with the merid- If this is positive, the angle is west of north; if negative, east ci/3 of north. »'' (9) Two trmms A and B start from the same point and A runs 60 miles northand 50 miles northeast. Find the displacement of B relative to A. Ans. 43.104 miles in a direction 34° 52' south of east. ' (11) A point undergoes two component displacements, 100 feet W. 30° S. and 30 ft. N. What is the resultant f / Ans. 88.88 ft. 13° south of west. -> (12) Three component displacements have magnitiides represented by 1,2 and 3 and directions given by the sides of an equilateral tri- angle, taken the same way around. Find the magnitude of the re- sultant. , Ans. 4/3. "p^ Two railway trains run on parallel roads, one 5 miles north, thS other Q miles south. Find the displacement of the last relative to the first. Ans. 11 miles south. (14) Two railway trains run, the one northeast a distance d, the other southeast the same distance. Find the displacement of the first relative to the last. Ans. d 4/2] direction north. )/ (15) A point undergoes three component displacements, N. 60° E. 40 ft., S. 50 ft , W. 30° N. GO ft. Find the resultant. wAns. 10 i''3 ft. W. "Sl^ "^ ^^^ ^^^9(^^^^ -^C of a quadrilateral ABCD is produced to E, J^^Jiat CE is equal to AC, show that AE is the resultant in the difkctior AC of the displacements AD, DB, BC and AC. \ V wl7) ABCD is parallelogram. E is the middle point of AB. FinxL y. CHAP. III.] DISPLACEMENTS — EXAMPLES. 41 the components^ in the directions of AB and AD, of a displacement which has the direction and half the magnitude of the resultant of component displacements represented by AC and AD. Ans. AE and AB. (18) Prove that the resultant of two equal displacements of mag- nitude a, inclined 60% is equal to the resultant of a and 2a inclined 120°. /'' : . ,- (19) Prove that if two component displacements are represented f' by'iwo chords of a circle drawn from a point P in the circumference perpendicular to each other, the resultant is represented by the diame- ', ter through the point. ^j (20) To an observer in a balloon the starting -ppint bears N. 20° E.,wjt JAand is depressed 30 below the horizontal. A point at the same level as the starting-point and 10 miles from it is vertically below him. ^j. What are the component displacements of the balloon with reference to the starting-point f Ans. 9.39 miles south, 3.43 miles west, and 5.77 miles up. v-J rA^^ \jJ ( 21 \ A point has the component displacements in the same plane, ' q~= 'iOft., d-i = 50 ft., da = 60 ft., making the angles with X and Y, a, = 60°, /? = 150° ; a^ = 120°, /^^ = 30° ; a, = 120°, ^a = 150°. Find the resultant displacement. Ans. (?x = 4- 20 - 25 - 30 = - 35 ft.; dy^- 34.64 + 43.3 - 51.96 = - 43.3 ft.; dr = V^x' + dy^ = 55.67 ft. cos a = ^ = g^, or a = 128° 57' 17" ; cos 6 = ^ =^^. OTb = 141° 3' 43". dr 55.67 dr 55.67 (22) A point has the component displacements d, = 40 ft., d, = 50 ft., da = QOft., making with X, Y, Z, the angles a, = 60°, ft, = 100°, r, obtuse ; a^ = 100°, fh — 60°, y^ acute ; a^ = 120°, A = 100°, y^ acute. Find the resultant displacement. Ans. We find the angles y, (page 12) by the equation cos' y = — cos (a 4- 0) cos (a — ft). Hence yi = 148° 2' 31".7, y-, = 31° 57' 28".3, y^ ^ 31° 57' 38".3. (Z^ = -f 30 - 8.6824 - 30 = - 18.6824 ft. dy =- 6.946 + 25 - 10.419 = + 7.635 ft. dz =- 33.937 + 42.431 + 50.907 = + 59.391 ft. -1 Q AQ04, dr = VdJ+'dy^ + d,» = 63.73 ft. cos a = — gg^^g— . or a = 107° 19* 36"; c,s6 = i|:|-, or 6 = 83° 0' 33"; cos c = "tg|^, or c = 18° 46' 43'. ^ fy) ^loy^ :v A CHAPTEE IV. VELOCITY. RESOLUTION AND COMPOSITION OF VELOCITIES. BKCTANQULAK COMPOKENTS OP VELOCITY. SIGN OP COMPONENTS OP VELOCITT. ANAIiTTIC DETERMINATION OP RESULTANT VELOCITY. Mean Velocity. — The distance described by a moving point per unit of time we have called the mean speed of the point (page 15). The displacement (page 34) of a point per unit of time we call the mean linear velocity. Therefore the number of units in the dis- placement of a point in any given time, divided by the number of units in that time, gives the number of units of mean linear velocity — or the magnitude of the mean linear velocity — while the direction of that velocity is the same as that of the displacement. The term velocity always signifies linear velocity imless other- wise specified. Mean speed, then, is mean time-rate of distance described (page 15). Mean velocity is mean time-rate of displacement. Thus if a point moves in any path P^AP^ from the initial position Pi to the position P2 in the time t Py^ — -p>P2 units, the magnitude of the mean speed is given by f^^ £■ —^ ?, while the magnitude of the mean ve- ''i 1 -A • ^- u displacement PjP^ ■, .. ,. locity IS given by - — ^ — -, and its direc- tion by the direction of P^P-t. If a point moves with uniform speed in a circle of radius r units and the 27tr time of revolution is t units, the mean speed is —7- units of speed. But the mean velocity in the time of one revolution is zero, because the disi)lacement TtT in that time is zero. Again, the mean speed in one half a revolution is — or 2' units of speed, as before, but the mean velocity for the same time has for its magnitude 77 or — units of velocity, and the direction is that of a ^iameter ■5' * from the initial to the final position of the point. Instantaneous Velocity. — The limiting magnitude of the mean speed when the interval of time is indefiiiitely small we have called 42 CHAP. IV.] VELOCITT. 43 (page 15) the instantaneous speed. In the same "way, the limiting magnitude and direction of the mean velocity when the interval of time is indefinitely small is called the instantaneous velocity. The terms speed and velocity should always be understood to mean instantaneous speed and instantaneous velocity unless other- wise specified. Using- the terms thus, we see that when t is in- definitely small, distance described and displacement coincide and hence the speed at Pi is the magni- tude of the velocity at P, , while the direction of this velocity at any instant is that of the tangent to the Pi path at that instant. Velocity is directed speed. Speed is magni- tude of velocity. Unit of Velocity. — Since, then, the magnitude of the velocity at any instant is the speed in the direction of the velocity at that in- stant, it follows that the unit of velocity is the same as the unit of speed (page 15j, or one unit of length per vmit of time, as, for instance, one foot per second. For this reason we have used (page 25) the letter v for the niimeric for speed. Uniform Velocity. — When the velocity has the same magnitude and direction whatever the interval of time, it is uniform. Uniform velocity, then, is necessarily uniform speed in a straight line in a given direction. The velocity in such case is the same as the mean velocity for any interval of time. Variable Velocity. — When either the magnitude or direction of the velocity changes it is variable. When the magnitude alone changes, we have variable speed in a straight line. When the direction only changes, we have uniform speed in a curved line. When both change, we have variable speed in a curved Une. In all these cases the velocity is variable. Thus we can speak of a point moving in a circle with uniform speed, but we cannot speak of a point moving in any curve at all with uniform velocity. If the velocity is uniform, the path must be straight, as well as speed constant. A point can be projected with the same speed in many different directions, but we cannot speak of the same velocity in different directions. A change of direction is a change of velocity whether the speed changes or not. Velocity a Vector ftuantity.— Since velocity is speed directed, or time-rate of displacement, it has not only sign and magnitude like speed (page 15), but also direction like displacement, and is there- fore a vector quantity. Line Representative of Velocity.— Velocity, then, can be repre- sented like displacement (page 34) by a straight line. The length of this line represents the magnitude of the velocity, and the direc- tion as denoted by an arrow gives the direction. Thus the straight line A.Aj represents by its length the magnitude of a velocity, and the arrow shows its direction. Triangle and Polygon of Velocities. — The prin- ciples therefore of page 35 hold good for velocities as well as displacements. We have then the triangle and polygon of velocities. Resolution and Composition of Velocities.— We can also combine and resolve velocities in the same way precisely as displacements, and the principles of page 36 apply here also. 44 KINEMATICS — GENERAL PRINCIPLES. [CHAP. IV. ^ Eectangular Components of Velocity.— If a point moves in any path from P, to P in the time t, the dis- placement is the chord PiP, and the mean '" 7 V • chord P,P velocity 18 v^ — t — . ->vx If X, and Pi are the co-ordinates of P,, ^ and X and y of P, then the horizontal and vertical components of the mean velocity 4- Wi X V are c VI J)t t'% If the time is indefinitely short, we have in the notation of the Calculus, for the instantaneous velocities, dx dy nJL, I'P, ^Jio. rx = dt Vy = dt' Sign of Components of Velocity. — We see that if Vx is directed towards the right, we have in the 1st and 4th quadrants x numeric- ally greater than x, and both are positive. In the 2d and 3d quad- rants, if Vx is directed towards the right, x, is numerically greater than X and both are negative. In all quadrants, then, Vx will be positive when directed towards the right. In the 1st and 2d quad- rants, if Vy is directed upwards, y is numerically greater than yi and both are positive. In the 3d and 4th quadrants, if Vy is directed upwards, yt is numerically greater than y and both are negative. In all quadrants, then, v,/ will be positive when directed upwards. We have then the following general rule for the signs of the components of the velocity in any quadrant : If the direction of the line representative is toivards the right, Vx / *' positive ; if towards the left, Vx is negative. If upwards, Vy is positive; if downwards, negative. The sign then as applied to component velocities indicates direc- tion of motion. For rectangular co-ordinates (+) signifies towards the right or upward, (— ) towards the left or downward. For three rectangular axes OX, OY, OZ, let a point Pbe given by the space co-ordinates x, y, z. Let the velocity v of the point make the angles «, fi and r with the axes of X, Y and Z. Then we have / X Vx — V cos a, Vy =V cos /J, Vz=V COS y y\ /z .z for the components in the direction of the X axes. When the angles a^ (i^ y are acute or less than 90°, these components are posi- tive ; when the angles are obtuse or more than 90% these components are negative. Therefore, as before, Vx towards the right is positive, towards, the left negative, Vy upwards is positive and downwards negative, and Vz in the direction OZ is positive, in the opposite direction, negative. If polar co-ordinates are used, the component velocity along the radius vector must he taken as positive when it acts away from the pole, and negative when it acts toioards the pole.* * Evidently, then, we measure angles in the plane XYfrom OX around to- wards 07"; in the plane YZ from OF around towards OZ; in the plane ZX from OZ around towards OX, as shown by the arrows in the figure. •CHAP. IV.] VELOCITY — EXAMPLES. 45 Analytical Determination of the Resultant for Concurring Veloci- ties.— When the line representatives meet in a point they are dialled concurring. We have then the same expressions for the magnitude and direction of the resultant of any number of concur- ring component velocities as for displacements (pages 36 and 37). We have only to substitute v in place of d. EXAMPLES. (1) To a man driving easttvard with a speed of 4 miles an hour, th3 ivind blows apparently from, the north, hut ivhen he doubles his speed the wind appears to blow from the northeast. Find the real direction and velocity of the ivind. Ans. The wind blows from the northwest with a velocity of 4 1^2 miles an hour. *' (2) A point moves in a straight line from A to B, QOft. W. 30° S., in 10 sec, and thence in a straight line to C, 30 ft. N., in 20 sec. Find the mean speed and the mean velocity. Ans. The length of path is 90 ft. traversed in 30 sec, or mean speed is 3 ft. per sec. The displacement is 30 |/3 ft. W., or the mean velocity is V3 ft. per sec. W. "" ^ (3) A ship sails N. 30° E. at 10 miles an hour. Find its easterly velocity and its northerly velocity. Ans. 5 miles an hour; 54^3 miles an hour. (4) A river-ir^iii&ixxBXd has a current of 5 miles per hour, and <x boat capable of making 10 miles an hour in Stillwater is to go ■Straight across. In what direction m,ust the boat be steered f - >■< Ans. Up stream in a diiection inclined 60° with the bank. _ , (5) Find the vertical velocity of a train moving up a 1-per-cent 1^^ gradient at a speed of 30 miles per hour. /!?«£•♦ Ans. 0.3 mile per hour. ff /j (6) A man travelling 4 miles per hour east finds the wind to come from the southeast. When he stands still it shifts 5° to the south. Find its velocity. /o° Ans. 32.52 miles per hour N. 40° W. l^-'tf) A. point moving with uniform speed in a circle of radius 30 ft. describes a quadrant in 10 sec. Find the mean speed, the mean velocity, the instantaneous speed and the instantaneous velocity. TtT Ans. The length of path described in 10 sec. is -5- = 47.12 ft. The mean speed is then' 4. 712 ft. per sec, and since_this is uniform it is also the instan- taneous speed. The displacement is r |/2 = 42.42 ft. at an angle of 45° with the diameter through the starting-point. The mean velocity is then 4.242 ft. per sec. in the same direction. The instantaneous velocity is at any instant tangent to the circle at the point at that instant, and equal in magnitude to the instantaneous speed, or 4.712 ft. per sec. V \\aj«) J. man loalks at the rate of 4 miles per hour in a rain-stoi-m, ys<md the drops fall vertically tvith a ^peed of 200 ft. per sec. In what ' d/irection will they seem to him, to fall f Ans. Inclined 1° 40. 8 to the vertical. v/ 46 ^ KINEMATICS — GENERAL PRINCIPLES. [CHAP. IV. \.--^ A ship sails east with a speed of 12 miles an hour, and a shot is fired so as to strike an object which bears N.E. If the gun gives the shot a mean horizontal velocity of 90 ft. per sec, towards what point of the compass must it point f Ans. N. 37° 3' E. ~~ (10) A man 6 ft. tall ivalks at the rate of 4 miles per hour directly away from a lamp-post 10 ft. high. Find the velocity of the ex- tremity of his shadow. Ans. 10 miles per hour in the direction he is walking (see Ex. 14, page 19). (11) Two points moving with uniform speed v start at the same instant in the same direction from, the point of contact of their paths. The one moves in a circle of radius r, the other in a tangent to the circle. Find their relative velocity at the end of the time t. Ans. 2® sin k- in a direction inclined to the tangent at an angle -in 1 radians. •X'^i) A moves N.E. with a velocity v, and B moves S. 15° E. with the same velocity. Find A's velocity relative to B. Ans. V t^S, dxasHm^lStCS- ^^""^ (13) AxO'il^^y train runs 30 miles per hour north. Another running is miles per hour appears to a passenger in the first to be running at 25 mites per hour. Find the direction of the velocity of the latter. ^^ Ans. N. 56' 15 E. or N. 56' 15' W. (14) Two candles A and B, each 1 ft. long and requiring 4 and 6 hours respectively to burn out, stand vertically at a distance of I ft. The shadow of B falls on a vertical wall at a distance of 10 ft. from, B. Find the speed of the end of the shadow. Ans. 8 inches per hour. (15) A ship has a northeasterly velocity of 12 knots per hour Find the magnitude of her velocity {a) in an easterly direction; (6) in a direction 15° W. of N. Ans. (a) 6 ^^2 ; (b) 6 knots per hour. (16) A boat-crew row 3i miles down a river and back again in 1 hour 40 minutes. If the river has a current of 2 miles per hour, find the rate at which the creiv would row in stilt water. Ans. 5 miles per hour. v., (17) A steamer goes 9.6 miles per hour in still water. How long /V will it take to run 10 miles up a stream and return, the velocity of r the stream being 2 miles an hour f Ans. 2 hours 11 minutes. (18) A steam tug travels 10 miles an hour in still ivater, but draws a barge 4 miles an hour. It has to take the barge 10 miles tip ,'ya stream ivhich runs 1 mile an hour, and then return without the Jv barge. How long will it take for the job f A Ans. 4/5 hours. (19) A vessel makes two runs on a measured mile, one with the tide in a minutes and one against the tide in b minutes. Find the CHAP. IV.] VELOCITY — EXAMPLES. 47 velocity of the vessel through the water, and of the tide, supposing both uniform. Ans. 3Q- — — - and 30 — -- miles per hour. ^^ ab ah (20) A point receives simultaneously three velocities, 60 ft. per sec. N., 88 ft. per sec. W. 30° S., and 60 ft. per sec. E. 30" S. Chive the magnitude and direction of the resultant velocity. Aus^ feet per sec. W. 30° S. il) A ship sailing due north at the rate of 8 miles per hour is )^ carried to the east by a current of 4 vniles per hour. Find the ve- locity ivith reference to the land. Aus. j3.9'i miles per hour N. 26° 34' E. ^-t§2) A ship is sailing E. 22i° S. at the rate of 10 miles an hour and the wind seems to blow from the N. W. unth a velocity of 6 miles an hour. Find the actual velocity of the wind. Ans. 15.7 miles an hour W. 30° 55' N. (23) A point moves in t seconds from A to B, the positions being given by the co-ordinates Xi , y^ and Xt, y^. What is the mean ve- locity f V{Xi — Xiy-\-{y, pif^ making an angle a with the axis of x a -yi Ans. « = , given by tan a = - X2 — Xi ^^ (24) A point has four component velocities in the same plane, of 12, 24, 36, 48 ft. per sec, making with the axis of X the angles of 16", 29°, 33°, 75° respectively. What is the resultant velocity i Ans. Vx = 75.14; Vy = 80.915; Vr = 110.424; angle with axis of x, a = 47" 7' 10"; angle with axis of y, b = 42° 52' 50". C^ (25) A point has three component velocities in the same plane given by v, = 40, v-i = 50, Va = 60 ft. per sec., maMng with the axes of X and Y angles given by ^i = 60°, /3, obtuse; /?» = 30°, as obtuse; 0-3 = 120°, /Sa obtuse. Find the resultant velocity. Ans. We have /S, = 150°, a, = 120°, /?s = 150°. Hence ij^; = - 35 ft. per sec, Vy — — 43.3 ft. per sec. , Vr = 55.67 ft. per sec, making the angles with X and Fgiven hy a = 128° 57' 17", b = 141° 2' 43 '. l^ (26) A point has the component velocities u, = 40, Vi = 50, Vs = 60 ft. per sec., making the angles with X, Y, Z, a, — 60°, /?i = 100°, xi obtuse ; a^ = 100°, /?3 = 60°, yt acute ; as = 120°, /?3 = 100°, yz acute. Find the resultant velocity. Ans. We find the angles y (page 12) by the equation cos" y = — cos {a -\- fi) cos (a — jS). Hence y, = 148° 2' 31" 7. ya = 31° 57' 28 '.3, y^ = 31° 57' 28".3. Vx = — 18.6824 ft. per sec, % = + 7.635 ft. per sec. ,Vz = -\- 59.391 ft. per sec, Vr = 62.73 ft. per sec, making with the axes of X, T, Z, angles given by a = 107° 19' 36", b = 38° 0' 33", c = 18° 46' 42". pi Ak (f ^M^ CHAPTEK V. /IfV*- ACCELERATION. RESOLUTION AND COMPOSITION OF ACCELERATIONS. ANALYTICAIi DETERMINATION OF RESULTANT FOR CONCURRING ACCELERA- TIONS. EQUATIONS OP MOTION. THE HODOGRAPH. TANGENTIAL AND NORMAL ACCELERATION. Mean Acceleration.— Let P, , P2, etc., Fig. (a), be the path of a moving point P, and let the corresponding instantaneous velocities Fig. (a). Fig. (6). in the time i is a = be ^1,^2. etc Each velocity is tangent to the path at the corresponding point and is equal in magnitude to the speed at that point. If t is the number of units of time in passing from Pi to P2 , the mean speed for that time (page 15) is ^-^^-^i^ units of speed. The integral change of speed in the time t is t?2 — Vi (page 24), and the mean rate of change of speed V-i — v> t (page 24). If now in Fig. (6) we draw OQi parallel and equal in magnitude to Ui and OQi parallel and equal to Vi , then the integral change of speed is represented by OQi — OQi and the mean rate of change of speed by '^^-Qg' . The change of velocity, however, in the time t is represented in magnitude and direction by Qi Q^ , and this we call the integral ac- celeration. The mean time-rate of change of velocity is given in magnitude by ^y^\ and in direction by QiQi- We call this the mean linear acceleration. The term acceleration always means linear accelera- tion unless otherwise specified. Mean acceleration is then time-rate of change of velocity, whether that change takes place in the direction of motion or not. Instantaneous Acceleration. — The limiting magnitude and direc- tion of the mean acceleration, when the interval of time is indefi- nitely small, is the instantaneous acceleration. The limiting direction is not necessarily tangent to the path 48 €HAP. v.] ACCELERATION. 49 except in the case of rectilinear motion, and the limiting magnitude is not the rate of change of speed in the path, except in the case of rectilinear motion. The term acceleration always signifies instantaneous accelera- tion unless otherwise specified. It is the limiting time-rate of change of velocity, whether that change take place in the direction of the motion or not. Acceleration may he zero, uniform or variable. When it is zero the velocity is uniform, and we have uniform speed in a straight line. When it is uniform, it has the same magnitude and the same unchanged direction, whatever the interval of time. In such case the acceleration at any instant is equal to the mean acceleration for any interval of time. If its direction coincides with that of the velocity, we have uniform rate of change of speed in a straight line. If it does not so coincide, we have uniform acceleration and motion in a curved line. When it is variable, either direction or magnitude changes, or both change. Unit of Acceleration. — The magnitude of the imit of acceleration is evidently the same as that for rate of change of speed, viz., one unit of length-per-sec. per sec, as for instance one foot-per-sec. per sec. We denote the magnitude of the acceleration thus measured by the letter/, to distinguish it from rate of change of speed, which we have denoted by a (page 25). Line Representative of Acceleration. — Since acceleration is time- rate of change of velocity, and is therefore, like velocity and dis- placement, a vector quantity, it can be represented like them by a straight line, whose length and direction give the magnitude and direction of the acceleration (pages 34, 43). Rate of change of speed is given by stating sign and magnitude only. It is a scalar quantity (page 25). Triangle and Polygon of Accelerations. —The principles, there- fore, of page 35 hold good for accelerations also, as well as dis- placements. We have then the " triangle and polygon of accelera- tions." Composition and Resolution of Accelerations. — For the same reason we can covapine and resolve accelerations in the same way as displacements, and the principles of pages 35, 36 apply. Let OB and OD be the initial and final velocities of a point in any given time t. Then BD is the integral acceleration and TiD — is the mean acceleration, or the instan- taneous acceleration if it is uniform. If not uniform, ^^, where t is indefinitely small, gives the acceleration. Draw OA and DC at right angles to any line AO through B in any given direction. Then — ^ is the component of the acceleration in this direction. But — = ^^~^, and AC and AB are the components of the velocities in the direction AC. Hence the component in any direction of an acceleration is equal to the acceleration of the component velocities in that direction. 50 KINEMATICS — GENERAL PRINCIPLES. [CHAP. Y, Sign of Components of Acceleration. — We have the same rule for the signs of the horizontal and vertical components fx and fy of any acceleration/, as for the horizontal and vertical components of Vx and Vy of any velocity v (page 44). If the direction of the linear representative is towards the right or upwards, fx and fy are positive ; if towards the left or doivn- ward, fx and fy are negative. The sign, then, applied to component accelerations indicates di- rection of action. For rectangular co-ordi- nates (+) signifies in the direction OX, OY, OZ, (— ) in the opposite directions. If polar co-ordinates are used, the compo- nent acceleration along the radius vector is positive ( + ) when it acts away from the pole, and negative (— ) when it acts towards the pole. Evidently, then, we measure angles in the plane XY from OX around towards OY, in the plane YZ from OY around towards OZ, in the plane ZX from OZ around toward OX, as shown by the ar- rows in the figure. Take, for instance, the case of a particle projected vertically away from the earth with the initial velocity «, and attaining the final velocity v. As long as the particle ascends, the direction of v is upwards, and v, Vi are both positive. The acceleration due to gravity is always downwards, and hence is negative. When the particle is descending with the initial velocity Vi , then both Vi and V are negative, and the acceleration is negative as before. Analytical Determination of the Resultant for Concurring Accelerations. — When the line representatives meet in a point, they are said to be concurring. We have then the same expressions for the magnitude and direction of the resultant of any number of con- curring component accelerations as for displacements (pages 37 and 38). We have only to substitute/ in place of d. Equations of Motion of a Point under Diflferent Accelerations. — The magnitude of the acceleration in general is not the same as rate of change of speed, except in the case of rectilinear motion. We have therefore denoted it by /, to distinguish it from rate of change of speed, which we have denoted by a. If, then, we denote by ft the magnitude of the tangential acceleration, or the tangential component of /, we have ft = a, or the magnitude of the tangential acceleration is equal to the rate of change of speed. (a) Acceleration Zero. — We have in this ^ case the line representative QiQ'i = 0, and ■„^^'']<P hence the line representative of the velocity ^^"'^K^f^^^ does not change either in direction or magni- p-=i_ - > ^q tude. '' ^^ We have then rectilinear motion with constant velocity, and for any interval of time S — Si .^. ^=-r-' ^'> where s, and s are the initial and final distances of the point from any fixed point in the line. This equation is general if we pay attention to the sign of displacement and velocity (pages 37 and 44). (b) When the Direction of the Acceleration coincides with the- Direction of the Velocity. — In this case the line representative QiQt CHAP, v.] EQUATIONS OF MOTION. 51 coincides in direction with PQi — Vi. We have then rectilinear motion with varying velocity. If / is uniform, the instantaneous acceleration is equal to the mean acceleration for any interval of time. Hence, if u, is the initial and v the final velocity, we hsuveforf constant V — Vi f=—r- (2) This value of / is general when we pay attention to the sign for velocity and acceleration (pages 44 and 50). From (2; we have v = vi +ft (3) The average speed is V+ Vi 1 ^^ ~2~ ^^' + 2^* (4) The displacement is s-s,='^^^t = v^t + ^ft' (5) Inserting the value of t from (2) we have v" — v,^ s-s,= g^ ■ (6) Hence v" = Vi"" + 2fis — s.) (7) [Tffis variable, we have from (1), in Calculus notation, ds ^ = ^' («) and from (3), f_dv_^^ and from (8), s-8,- I tdt (10) ! — «1 = / « The preceding equations can be deduced from these as on page 28.] All these equations are precisely similar to those on page 28, except that we have / in place of a. (c) When the Acceleration is Constant in Magnitude and Direction, but makes an Angle (p with the Initial Velocity. — In this case we have motion in a curve, and QiQ-i =ft. Hence v" = (vi + ft cos 0)' + (ft sin (py = V,'' + 2vift cos <P + fH\ . (11) The tangent of the angle QiPQ^ is tan^.P^. = -^$^ (12) Vi+ft cos <t> The displacement in the time t is given by ip = Ut + ^/i^ cos (p\\ (|/«'sin 0J'; . . . . (13) ahd the tangent of its angle of inclination to Ui is iff sin ,j^. Vit + Iff' cos 53 KINEMATICS — GENERAL PKINCIPLES. [chap. V. The Hodograph. — Let a point P moving in any curve have the positions Pi , I\, Pz, etc., and let the magnitude of the corresponding velocities be Vi, zh, fs, etc. These velocities are tangent to the path at P, , P2, P3, and are equal in magni- tude to the speed at these points. If from a point O we draw lines OQ,, OQi, OQ3, etc., parallel and equal to v,, v^, Vs , the extremities of these lines will form a polygon. If, however, the points Pi, P2, P3 are indefinitely near, the polygon be- comes a curve ^1^2 ^3, such that when the point P describes the path P1P2P3, we can conceive another point Q to describe the curve QiQiQs. This curve is called the hodograph.* The point O is called the pole of the hodograph. The points Q,, Q2, Qs are called the points cor- Tesponding to Pi, P2, Pa. Any radius vector, as OQ,, OQ2, in the hodograph, represents in magnitude and direction the velocity at the corresponding point Pi, Ps, etc., of the path. The magnitude of OQi, OQn, etc., is the speed y., v-^, etc., at P, P,, etc. The direction of O^,, OQ^ is the direction of v,, Vi, or that of the tangent to the path at Pi , P2. If t is the interval of time in moving from P, to P2 , then the chord Q, Q^ in the hodograph gives the magnitude and direction of the integral acceleration for that time, and ^^Vi ^hich is the mean velocity in the hodograph, gives the mean acceleration in the path. When the time is indefinitely small, QiQi becomes tangent to the hodograph and ^'^'' , which is now the instantaneous velocity in the hodograph, gives the instantaneous acceleration in the path. The tangent to the hodograph at any point, therefore, gives the direction of the instantaneous acceleration at the corresponding point of the path. The instantaneous speed in the hodograph gives the magnitude of this acceleration. Hence, the velocity at any point of the hodograph is the accelera- tion at the corresponding point of the path. Tangential and Normal Acceleration.— The entire resultant ac- celeration / at any point of the path may be resolved into a com- ponent tangential to the path ft -and a component normal to the path/n, so that/= Vft" + fn'. The magnitude of the tangential -component ft is the rate of change of speed in the path. Its direction is always tangent to the path, or parallel to the radius vector of the hodograph at the corresponding point of the hodograph. In order to find the normal com- ponent fu , let us first suppose a point to move in a circle with con- stant speed v. * Invented by Sir W. R. Hamilton. CHAP, v.] TANGENTIAL AND NORMAL ACCELERATION. 53 Let the radius of the circle be r, and take any two points Pi and Pn on the circle. Then the velocity at Pi is v tangent at Pi or per- pendicular to r at that point, and the velocity at Ps is v tangent at Pa or perpendicular to r at A. Now construct the hodograph by making OQi parallel and equal to the velocity at P and O^a parallel and equal to the velocity at P2,etc. Evidently the hodograph is also a circle of radius v, and the speed in the hodogi'aph is also constant, since the point P moves with constant speed and makes a revolution in the same time as its corresponding point Q in the hodograph. Let t be the time of revo- 27TV lution ; then -^ is the speed of Q in the hodograph or the accelera- tion of P in the path ; and since this speed at any point Q^ is at right angles to OQi or v, it is normal to the path at P or parallel to CP. We have then /n = ^-. But the speed in the path is v — ——^ 2tiv Hence t = — , and substituting, we have V We can obtain the same result as follows : Since the point P" moves from Pi to Pt in the same time that Q moves from Qi to Q2 , and the angle P1CP3 is equal to the angle QiOQt, we have fn: V-.-.V.r, or fn= — • r Since we have supposed v constant in magnitude, the tangential acceleration ft is zero, and therefore fn in this case is the magnitude of the total resultant acceleration /. A normal acceleration, then, has no effect upon the speed, but only changes the direction of motion. Let us now suppose that the speed v is not constant, the point still moving in a circle. Then the hodograph will not be a circle. But if the two points P and Pa are indefinitely near, so that the- arc PiPi is indefinitely small, the velocities at Pi and P can be taken as equal still, and we shall still have - « v" fn. V-.-.V.r, or fn= r Again, if the point P moves in any curve whatever, a circle can always be described whose curvature is the same as that of the curve at any given point. The radius of this circle is the radius of curvature p at the point. In this case, then, we should have fn-.V-.-.V-.p, or fn——. Therefore, in general, whatever the path may be and whatever the speed in the path — The magnitude at any instant of the normal component of the acceleration is equal to the square of the speed at that instant divided by the radius of curvature. 54 KINEMATICS— GENERAL rEINCIPLES, [CHAP. V. if^^h aM-*^ "^ EXAMPLES. # ^ i (1) A point moves with uniform velocity in a straight line. What is the hoaograph f Ans. A point. ^ (2) A point moves with uniform acceleration either in a straight line or a curve. What is the hodograph f What is the speed in the hodograph f Ans. A straight line. Uniform. (3) Show that the direction of motion of any point B on the cir- cumference of a circle rolling with velocity v on a straight line is perpendicular to AB at any instant, when A is the point of contact of the circle with the straight line at the instant consider'ed. (4) AB is a diameter of a circle of which BC is a chord. When is the moment about A of a velocity represented by BC the greatest f Ans^^-When BC and AC are equal. U^) A point is moving with uniform speed of a mile in 2 min. 40 sec. round a ring of 100 ft. radius. Find the acceleration. An^^O.SQ ft.-per-sec. per sec. towards the centre. U^) A point moves in a horizontal circle ivith uniform speed v, starting from the north point and moving eastward. Find the integral acceleration when it has moved (a) through a quadrant ; (b) a semicircle ; (c) three quadrants. Ans. {a) V ^2^ SW. ; (6) 2v, W. ; (c) v ^2, NW. (7) If the component velocities of a moving point are represented by the sides of a plane polygon, taken the same way round, the algebraic sum of their moments about any point in their plane is zero. Ans. If the polygon closes, the resultant velocity is zero. If it does not close, the line necessary to make it close taken the other way round is the re- sultant. In either case the algebraic sum of the moments is zero. (8) Show that the hodograph of a point moving with uniform speed in a circle is a circle in which the corresponding point moves also with uniform speed. (9) Show that the locus of the extremity of a straight line repre- senting either of the tivo equal components of a given acceleration is a straight line perpendicular to the straight line representing the given acceleration and through its middle point. (10) Let the velocity of a point moving in a straight line vary as the square root of the product of its distances from two fixed points in the line. Show thai its instantaneous acceleration varies as the mean of its distances from the fixed points. Ans. Let « be one distance and a -f- s the other. Then from page 51, Chap. V, v — — = k i^s{a -[- s), where k is a, constant. _dv__ kads + 2ksds _ k^a + ^k^s _ k'^{a -f 2«) ^^"'~di~2dt y^T^TsT ~ 2 "" 2 ' But the mean of the distances is — —^ — — - = — - — . # ^ ^ i<t4r' CHAP, v.] ACCELEEATIOK — EXAMPLES. 55 (11) If the algebraic sum of the moments of the component ve- locities of a moving point about any two points P and Q are each zero, shoto that the algebraic sum of their moments about any point in the line PQ ivill he zero. (12) A moving point P has tivo component velocities one of which is double the other. The moment of the smaller about a point O in the plane is double that of the greater. Find the magnitude and di- rection of the resultant velocity. Ans. If a is the smaller component and a, /? are the inclinations of the gre ater and smalle r components respectively to PO, the resultant is a i/b -\- 4 cos {P -\- a), and it is inclined to PO at an angle whose sin is 2 sin a -f- sin /H i/5+4cos(/i4-aj (13) The velocity v of a point moving in a straight line varies as the square root of its distance sfrom a fixed point in the line. What is its instantaneous acceleration f (is — Ans. We have « = — - = A; Vs. where A is a constant. Hence at _dv _ kds _ kv _ B* _ A;' [4) Tivo raihcay trains move in directions inclined 60°. !Z%e one, A, is increasing its speed at the rate of ift.-per-min. per min. The other, B, has the brakes on and is losing speed at the rate of 8 ft.-per-min. per min. Find the relative acceleration. Ans. 4 1/7 ft. per min. per min., inclined to the direction of A at an angle whose sin is |/f and to the direction of B at an angle whose sin is ^ y'f. •""(IH) TTie initial and final velocities of a moving point during an interval of tivo hours are 8 miles per hour E. 30° N. and 4 miles per hour N. Find the integral and the mean acceleration. Ans. 4 4/8 miles per hour W., 2 V'3 miles-per-hour per hour W. HIC^Ta point moves in a circle of radius 8 inches and has at a given position a speed of 4 in. per sec, which is changing at the rate of 6 in.-per-sec. per sec. Find (a) the tangential acceleration; (b) the normal acceleration; (c) the resultant acceleration. Ans. (a) 6 in.-per-sec. per sec; (b) 2 in.-per-sec. per sec.; (c) 2 ^10 in.-per- sec. per sec. (17) Neivton assumed that the acceleration of gravity varied in- versely as the square of the distance from the earth's centre. He then tested this assumption by applying it to the moon. Assuming the acceleration at tlie earth's surface 32.2 ft. -per -sec. per sec, the radius of the earth 4000 miles, the distance betiveen centres of earth and, moon 240,000 miles, and the speed of the moon in its orbit <i round the earth 3375 ft. per sec, shoiv that Newton's assumption is in accord IV ith fact. Ans The acceleration of the moon's centre towards the earth is — , or r = 0.(X)89 ft.-per-sec. per sec. 240000 X 5280 But according to Newton's assumption, if g' is the acceleration at the dis- tance of the moon, ^ = -^^^ = 3^. Hence g' = 0.0089 ft.-per-sec per sec. * * 56 KINEMATICS— GENERAL PRINCIPLES. [CHAP. V, (18) Find the resultant of four component accelerations repre- sented by lines drawn from any point P within a parallelogram to the angular points. Ans. If G is the intersection of the diagonals, PC represents tlie direction of the reetfTtant, and 4PC its magnitude. (19) A ball is let fall in an elevator which is rising with an ac- celeration of 7.2 kilometers-per-min. per min. The acceleration of the ball relative to the earth is 981 cm.-per-sec. per sec. Find its ac- celeration relative to the elevator. Ans. 1181 cm.-per-sec. per sec. towards the floor. (20) Assuming the mean radius of the earth 6370900 meters, the speed of a point on the equator 465.1 m. per sec, acceleration of a falling body 9.81 m.-per-sec. per sec, find with what velocity a shot must be fired at the eqtiator with either a westerly or easterly direc- tion in order that, if unresisted, it shall move horizontally round the earth, completing its circuit in 1^ or li hours respectively. Ans. Westerly velocity, 8370.7 meters per sec. ; easterly velocity, 7440.5 meters^ef sec. ''^1) If different points describe different circles with uniform speeds and with accelerations proportional to the radii of their paths, show that their periodic times will be the same. (22) The resultant of two accelerations a and a' at right angles to one another is R. If a is increased by 9 units and a' by 5 units, the magnitude of R becomes three times its former value, and its direc- tion becomes inclined to a at the angle of its former inclination to a'. Find a, a' and R. Ans. 3, 4 and 5 units respectively. (23) If a tangent be drawn at any point of a conic section, the locus of the foot of the perpendicular let fall from a focus on this tangent is a circle in the case of the ellipse and hyperbola, and a straight line in the case of a parabola. Also the locus of the foot of a perpendicular from the vertex of a parabola on a straight line drawn through the focus is a circle. Assuming these properties, show that if a point move in either a circle, ellipse, hyperbola or parabola, so that the moment of its ve- locity about a focus is constant, the hodograph is a circle. (24) Show that if a point moves in an ellipse so that the moment of its velocity about the centre is constant, the hodograph is an ellipse. [Note that the area of the parallelogram formed by drawing tangents to an ellipse at the extremities of a pair of conjugate diameters is constant] (25) A bullet is fired in a direction towards a second bullet which is let fall at the same instant. Show that the line joining them icill move parallel to itself and that the bullets will meet. (26) Determine whether any of the following equations are pos- sible or not : (1) lOavst + Su's = ZgH* ; (2) vH — 4as -I- 3a = ; (3) 6v + 2g'asH' = Sa^'st*. Ans. The first gives us !:— j^ in each term and is therefore possible. The second gives us ^tj—, ^—^ and j-iy?rj, or each term refers to different kinds of CHAP, v.] ACCELERATIOIf — EXAMPLES. 57 quantities, and the equation is nonsense on its face. The third gives us ~^, [i]" and [LY, and is also nonsense. (27) A point moves in a straight line so that the number of units of distance sfrom the origin at the end of any number of seconds t 5 8 5 is given by s = 2 + -t -] — P + -<*. Find (a) the number of units of distance from the origin at the start ; (b) the velocity v at any in- stant ; (c) the acceleration a at any instant ; (d) the velocity at the start ; (e) the acceleration at the start. 5 3 15 3 15 Ans. (a) 2 units of distance; {b) v = -^-\- -xt-^ -^f^ ; (c) « = „ + j^J 5 3 (d) - units of distance per sec. ; {e) ^ units of distance-per-sec. per sec. (28) A point moves in a straight line so that the number of units of acceleration a at the end of any number of seconds t is given by a = 7 — —t -¥ 2<^ If Vi is the number of units of velocity at the o start, and Si the number of units of distance from the origin at the start, find the velocity anid the distance from, the origin at any in- stant. Ans. v = Vi-\-lt-^t''-\- y? ; (29) A point moves in a straight line so that the number of units of velocity v at the end of any nuniber of seconds t is given by 3 5 V = 5 — -t + -t^. Find the acceleration a and the distance s, if Si is 2 6 the initial distance. Ans. a = _ - ^- -«; (30) A point has three component accelerations in the same plane given by f\ = 40, /a = 50, fa = 60ft.-per-sec. per sec, making with the axes of X and Y angles given by oci = 60°, y3i obtuse; 02 = 30% «* obtuse; as = 120°, /?3 obtuse. Find the resultant acceleration. Ans. We have /?i = 150°, a^ = 120°, ^3 = 150°. Hence fx = — 35 ft.-per-§ec. per sec. ; /»/ = — 43.3 ft.-per-sec. per sec, and fr — 55.67 ft.-per-sec. per sec, making the angles with X and Y given by a = 128° 57' 17", b = 141° 2' 43". (31) A point has three component accelerations, f, = 40, fi = 50, fs = QO ft.-per-sec. per sec, making with the axes of X, Y, Z angles given by a, = 60°, ^, = 100', y, obtuse ; a-i = 100% (ii = 60°, y^ acute; as — 120°, hi = 100°, yi acute. Find the resultant acceleration. Ans. We find the angles y (page 12) by the equation cos" y = — cos (a -\- (i) cos (a — /8). 58 KINEMATICS — GENERAL PRINCIPLES. [CHAP. V, Hence yx = 148° 2' 31".7, y^ = 31° 57' 28".3, y^ = 31° 57' 28".3 ; fx=- — 18.6824 ft.-per-sec. per sec, fy = -\- 7.635 ft.-per-sec. per sec., fz =-\- 59.391 ft.-per-sec. per sec, fr = 62.73 ft.-per-sec. per sec, making with the axes of X, Y, Z angles given by a = 07° 19' 36 '. 6 = 83° 0' 33", c = 18° 46' 42". (32) Investigate the motion of a point whose initial velocity in a horizontal direction is 0, and in a vertical direction — 32 ft. per sec. The horizontal acceleration is fx= + 16 ft.-per-sec. per sec., and the vertical acceleration fy = + At ft.-per-sec. per sec. Ans. The resultant acceleration is (page 51) ajid it makes an angle A with the horizontal whose cosine is cos A = 4Uid an angle pi with the vertical whose cosine is cos yU 4/(16)" + (4<)* The horizontal velocity at the end of any time is vx = 16«. The vertical velocity at the end of any time is Cj, = _ 32 + 2t\ The resultant velocity is V = Vil^ty + (2t^ - 32)2 = 3«» 4. 3S^ imd it makes an angle a with the horizontal whose cosine is 8t «««« = , -^+16' and an angle fi with the vertical whose cosine is ^°^^ = ^-l6- The distance s described in any time is «= |<3+32«. The tangential acceleration is ft = a =fx cos a +/vCOJ The normal acceleration is (page 52) fi = a=f. cos a +/„cos /? = 16 X ^T^ + 4< X ^q:^ = 4t "The radius of curvature is (page 53) g» (2<'4-32)* ^~fn~ 16 • The horizontal distance described is X = 8t\ CHAP, v.] ACCELERATION — EXAMPLES. 59 The vertical distance described is y = |<^ - S2t. Eliminating t, we have for the equation of the path (33) Investigate the motion of a point whose initial velocity in the direction of the axis of x is + 2 ft. per sec.; in the direction of the axis of y, 0] in the direction of the axis of z, +4 ft. per sec. The acceleration in the direction of the axis of xisfx = 0; in the di- rection of the axis of y,fy= + Sft.-per-sec. per sec.; in the direction of the axis of z,fz= + 5 ft.-per-sec. per sec. Ans. The resultant acceleration is (page 51) /= -^/0'' + 3'' + 5''= i/34, which makes an angle A. with the axis of x whose cosine is cos A = — -= = ; i/34 with tho axis of y an angle /i whose cosine is 8 cos JJ with the axis of g an angle v whose cosine is 5 cos V = r=r. i/34 The velocities in the direction of x, y and z are Vx = 3; ?)y = 3<; «« = 4 + Sit The resultant velocity is « = |/2!> + {^tf + (4 + Uf = V'34«» + 40« + 20, -which makes angles a, /3 and y with the axes of x, y and z given by 2 ,, m cos p 4/34«« + 40« + 20 V^f + 40« + 30 4 + 5« cos y >^34i» + 40« + 20 The distances described in tho direction of x, y and z are x^2t; y = |«'; 8 = 4« + |<«. If we eliminate t, we have y=-^; 2 = 2a; + -5<B». o These are the equations of the projection of the path upon the planes of vy and OSS. (0 ^^y^ r^ CHAPTEE VL MOMENT OF DISPLACEMENT, VELOCETY, ACCELERATION. LINE REPRESENTATIVE. KESOLUTION AND COMPOSITION OF MOMENTS OF DISPLACEMENT, VELOCITY, ACCELERATION. ANALYTICAL DETERMINATION OP RESULTANT VELOCITY AND MOMENT FOR CONCURRING VELOCITIES, ACCELERATIONS AND DIS- PLACEMENTS. Moment of Displacement, Velocity or Acceleration about a Given Point. — The product of the magnitude of a displacement, velocity or acceleration by the magnitude of the perpendicular let fall from any given point upon its direction gives the magnitude of the moment of the displacement, velocity or acceleration with reference to that point. The perpendicular is called the lever arm, and the point is the centre of moments. Thus if AB is the line representative of any displacement, ve- locity or acceleration, and the perpendicular ? from any point O is p, the moment relative to O / is ± AB X p. If AB is a displacement d, we / have ± dp. If it is a velocity v, we have ± vp. / ^^ \ If it is an acceleration/, we have ± fp. l^,"'^^ / 1+ The (+) sign indicates that the radius vector o*^ 7A I moves around O counter-clockwise as in the ^\y figure. The (— ) sign denotes clockwise rota- tion. If we draw a line 00' through O parallel to AB, it is evident that the moment of AB is the same relative to any point in this line. Hence, so far as the moment is concerned, a displacement, velocity or acceleration may be considered as laid off from any point in its line of direction. Moment of Displacement, Velocity or Acceleration about a Given Axis. — Let OF be a given axis and y AB any displacement, velocity or ac- celeration. Take any plane XZ perpendicular to the axis OY, intersecting this axis in O. Project AB upon this plane in A!B'. Now AB can be resolved at A into a component parallel to AB' and a component perpendicular to A!B' and therefore parallel to the axis OY. The component parallel to OY causes no rotation about Y. The . moment of AB about the axis is then the moment of A!B' aboutthe intersection O of the plane XZ with the axis. CHAP. VI.] MOMENT OF DISPLACEMENT. 6^ Hence the moment of a displacement, velocity or acceleration about a given axis is the moment of the component in a plane per- pendicular to the axis about the point of intersection of that plane with the axis. Here again, so far as moment about any axis is concerned a displacement, velocity or acceleration may be taken as laid off from any point in its line of direction. Moment of Displacement.— If AB = d is a displacement, we have then for its moment about any point O, il/ = ± dp, and dp is twice the area of the triangle AOB. Hence the moment of a displacement about any point is twice the area swept through by tlie radius vector from that point. The unit of moment of a displace- ment is then the square of the unit of p\ length, as one square foot. \ /' If we draw 00' parallel to AB and V' erect a perpendicular DO' at the middle point of AB, then DO' =p and the moment about O is equal to the moment about any point in the line 0(7. If O'A — O'B = r, and the angle AO'B = 9, then we have for the displacement d d = 2r sin ^. Moment of Velocity — Acceleration— If AB = u is a velocity, we have for its moment about any point O, M= ± vp. Hence the moment of a velocity about any point is tvnce the areal velocity of the radius vector. The unit of moment of a velocity is then the square of the unit of length per unit of time, as one square foot per second. If AB = /■ is an acceleration, we have for its moment about any point O, M— ±fp. Hence the moment of acceleration about any point is twice the areal acceleration of the radius vector. The unit of moment of an acceleration is then the square of the unit of length per unit of time squared, or one square ft.-per-sec. per sec. Line Representative of Moment of Displacement — Velocity — Ac- celeration. — Since moment of displacement, velocity, acceleration, has thus both magnitude and direction, it is a vector quantity like displacement, velocity, acceleration themselves, and like them can be completely represented by a straight line. Thus the length of a line represents the magnitude of the moment dp, vp or fp. This line is always taken perpendicular to the plane of rotation of the radius vector, which is therefore known when the direction of the line representative is known. Finally we denote the direction of rotation in the plane by placing an arrow- head on the line, so that wJien ive look along the line in the direction of this arrow, the rotation is always seen clockwise. s Thus the line AB denotes by its length the magni- tvide of the moment dp, vp or fp. The plane of rotation of the radius vector is at right angles to this line, which is therefore coincident with the axis of rotation. The direction of rotation is clockwise in this plane when ^ we look from A to B. 62 i£INEMATICS— GENEEAL PRIKCIPLE8. [CHAP. VI.. Resolution and Composition of Moments. — The principles, there- fore, of pages 35 and 36 hold good for moments of displacements, velocities and accelerations, as well as for displacements, velocities and accelerations themselves. We have then the triangle and polygon of moments. Sign of Components of Moments. — The signs of the line repre- sentatives of the components along the axes of X, F, Z of a moment of displacement, velocity or acceleration foliow the same rule as for components of displacement, velocity or acceleration (pages 36, 44, 50). Hence components in the directions OX, OY, OZ are positive ( + ), in the opposite direction negative (— ). If then we look along the line representatives of the compo- nents towards the origin O, the rotation is always seen counter-clockwise. Therefore rotation from X towards F, F towards Z, Z towards JTis positive, in the opposite direc- tions negative. For polar co-ordinates, directions away from the. pole are posi- tive, towards the pole negative.* The algebraic sum of the moments of any number of component displacements, velocities or accelerations, about any point in their plane, or about any axis, is equal to the moment of the resultant displacement, velocity or acceleration about that point or axis. Let AB, AC represent two component displace- ments, velocities or acceler- ations of a point A. Then the resultant is AR. Let O be any point in the plane of the components either out- side or inside the angle be- tween the resultant and either component. Then in the first case area OAR = area OAB + area BAR — area ROB, and in the second case area OAR = area OAB + area ROB — area BAR. In both cases area BAR = area ROB + area OAC, since all three triangles have equal base BR, and the altitude o!: BAR is the sum of the altitudes of ROB and OAC. We have then in the first case area OAR = area OAB + area OAC, and in the second case area OAR = area OAB — area OAC. * Evidently, then, we measure angles in the plane X I^ from OX around towards OF ; in the plane YZ from OT around toMip,rds OZ ; in the plane ZX from OZ around towards OX, as shown by the arrows in the figure. CHAP. VI.] RESOLUTION AND COMPOSITION OF MOMENTS. 63. But twice the area OAR is the moment of the resultant AH, and twice the areas OAB and OAC are the moments of the componenta AB and AC about O. Hence the moment of the resultant is equal to the algebraic sum of the moments of the components. If we have a third displacement, velocity or acceleration at A, the resultant of this and AR would be the resultant for all three. Hence the principle holds for any number of components. Again, let AB, BC, CD represent the components of a point A. Then the resultant is AD. Let OY be an axis and XZ a plane perpendicular to the axis. Let ab, be, cd be the projec- tions on this plane of the component velocities. We have just proved that the mo- ment of ad about O is the algebraic sum of the moments of the compo- nents ab, be, cd. But the moment of each of these " about O is the moment of AB, BC, CD about the axis (page 60). Hence the moment of the resultant AD about the axis is equal to the algebraic sum of the moments of the components AB, BC, CD. The moment of acceleration of a moving point relative to any fixed point in the plane of its motion is equal to the time-rate of change of the moment of its velocity about the same point. Let AB = t;i be the instantaneous velocity of a point A and f its instantaneous acceleration. Then in any indefinitely small time dt the change of velocity is BC =fdt, and the resultant velocity is .AC = -y in the plane of y, and /. Take a point O in the same plane and drop the per- pendiculars li , I and p upon the directions of Vi , V and/. Then, since the moment of the resultant is equal to the algebraic sum of the moments of the components, we have vl = Vili + fdt . p, or fp Vl — Vili di • If the path is a circle of radius r, then l = li =r, and we have relative to the centre fp=ftr. We obtain the same result as follows : Resolve the acceleration / into components ft tangent to the circle and fn normal. The latter component passes through the centre, and its moment is zero. We have then the moment of / Y equal to the moment Vy Vr Pi *^ " \ of the other ponent, or fp=ftr. General Analytical Determination of Resultant Velocity and Moment for Any Number of Concurring Com- ponent Velocities in the Same Plane. —Let the point P be given by the co-ordinates x, y. Let the component velocities of P be tJi , Ua , v» » 64 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VI. «tc., all in the same plane XY and making with the axes of X and Fthe angles cxi , /J, ; a., , /ia ; tf^s , /is , etc. Let V be the resultant velocity making the angles a and & with the axes of X and i' respectively. For the component velocities parallel to X and Y we have Vx = V, cos a, + V2 cos cr^ + Vi COS CVa + . . . = ^UCOS «; \ Vy = Vi COS /^j + Vi COS /?2 + ?'3 COS pa + . . . = 2v COS /i. f In these summations components towards the right or upwards are positive, towards the left or downwards negative. The resultant velocity is then Vr = Vvx' + Vy\ (2) making with the axes of X and Y angles a and 6 given by cos a = — , cos b = ~: (3) Vr Vr The moment of the resultant velocity with reference to O is the algebraic sum of the moments of the component velocities Vx and Vy. If p is the lever arm of v,. , we have, paying regard to the sign for du'ection of rotation, for the moment Mz about the axis OZ, or the moment in the plane XY, Mz = VrP=VyX — Vxy (4) Hence the lever-arm is Mz _, p = — (5) ^ Vr In these equations Vx, Vi,, x and y are positive in the directions OX, OF and negative in the opposite directions. With these con- ventions the equations are general. If Mz comes out positive, the direction of rotation about O is counter-clockwise; if negative, the direction of rotation is clock- wise as shown in the figure. In the first case the line representative is positive and therefore passes through O in the direction OZ. In the second case it has the contrary direction. In both cases, if we look along the line rep- resentative towards the origin, the direction of rotation will be seen as counter-clockwise. Since Vr may be considered, so far as the moment at O is con- cerned, as acting at any point in its line of direction (page 60), let us take it acting at E, the intersection of the line of direction of V with the axis of Y. Then we have for the distance OE, Mz Vx X OE = — Mz, or OE = . The tangent of the angle which Vti Vr makes with the axis of X is -— . Hence the equation of the line Vx of direction of the resultant velocity v is Vy Mz ,„. y = -x — — (6) Vx Vx If in this equation we make x = 0, we find the ordinate of the point in which the direction of the resultant velocity Ur intersects the axis of Y, viz., OE = y'=-'^ (7) Vx <:HAP. VI.] MOMENT OF VELOCITY. 65 If we make y = 0, we find the abscissa of the point in which the -dii-ection of the resultant velocity v,- intersects the axis of X, viz., X = — (8) General Analytic Determination of Resultant Velocity and Mo- ment for Concurring Component Velocities not in the Same Plane. — Let the point P be given by the space co-ordinates x, y, z, T x . .. and let the component velocities of P be y. , ^2, fa, etc., making with the axes of X, Y, and Zthe *-m angles ^r, , /i, , :ki ; as , /ij , /'a ; as, /-^3, ya, etc. Let i?rbethe resultant velocity making the angles a, 6, c with the axes. We have then for the compo- nent velocities parallel to X, Y. z ^^^-'T^ and Z -^^My Vx = Vi COS Oil + Vi cos aa + Vi COS as = ^V COS oc ; ^ Vy = V, COS /3, + V'2 COS A + Us COS /?3 = 2VC08/3; ( . . (1) Vz = u, cos^i + ??2Cosr2 + Vs COST'S = ^vcosy. ) In these summations components in the directions OX, OY, OZ are positive, in the opposite directions negative. The resultant velocity is then "/ Vr = I Vx" + Vj," + t?2*, (2) making with the axes of X, Fand Z angles a, 6 and c given by cosa=-, cos 6= J'-, cosc=-- (3) Vr Vr Vr The moment of the resultant velocity Vr with reference to O is the algebraic sum of the moments of the component velocities Vx , Vy and Vz . We take the positive direction of rotation in each of the co- ordinate planes in the direction indicated by the arrows in the figure. Thus, rotation about Z from Xto Y') " " X " F " Z >• are positive ; " Y " Z '' X) in the contrary directions, negative. We have then for the moments about the axes moment about Z parallel to plane XY, Mz = VyX — Vxy; ) " X " " " YZ Mx = vzy-Vyz-A. . . (4) " Y " " " ZX, M„ = VxZ - VmX. ) In these equations Vx , Vy , Vz and x, y, z are positive in the direc- tions OX, OY, OZ, and negative in the opposite directions. With these conventions the equations are general; and if Mz Mx, My come out positive, we have rotation in each plane counter-clock- wise as indicated by the figure ; if negative, clockwise. In the first case the line representatives pass through O and have the directions OZ, OX, OY. In the second case they have opposite directions through O. In any case the direction of rota- 66 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VI. tion is always counter-clockwise when we look along the line repre- sentative towards O. The equations of the projections of the line of direction of the resultant velocity v,- upon the co-ordinate planes are found as in the preceding Article, since each may be considered as acting at any point in its line of direction : Vy Mz on plane XY, y = -^-x : Vx Vx " " YZ,z = -^y-^: Vy Vy My_ Vz ' ZX, X = ^z Vz (5) If in these equations we make 2; = 0, we find the co-ordinates of the point in which the direction of the resultant velocity Vr pierces, the plane XY, viz., 0!' = - Vz' ^ Vz (6) If we make x = 0, we have for the co-ordinates of the point where it pierces the plane YZ, My z = y = Vx Vx (7) If we make 2/ = 0, we have for the co-ordinates of the point where it pierces the plane ZX, z' = -^, x' = ^ (8) Vy Vy Combining the line representatives of the moments given by (4) we have Mr = VrP = VMx' + My' + Mz' (9> Hence Mr P = Vr (io> The line representative for the resultant moment Mr passes through O and makes the angles d, e, f with the axes of X, F, Z given by cos d = Mx Mr' COS e = _My_ Mr' cos/ Mz Mr (11) Looking along this line representative towards O, the direction of rotation is always counter-clockwise. The projections of this line representative upon the co-ordinate planes make angles with the axes given as follows: M projection on XY tangent of angle with X= j^- Mx YZ ZX Y=^ (12) Z^ M, If we make Vz = 0, we obtain the equations of the preceding CHAP. VI.] ACCELBEATION — EXAMPLES. 67 article. If we make x,y, z zero, we have the equations of page 37 if we put V in place of d. General Analytic Determination of Resultant Acceleration and Moment for Concurring Component Accelerations. — The equations of the last two Articles hold good for a point having component accel- erations as well as for component velocities. We have only to sub- stitute / in place of v. General Analytic Determination of Resultant Displacement and Moment for Concurring Component Displacements. — The same equa- tions hold for a point having component displacements. We have only to substitute d in place of v. If we then make x, y, z = we have the equations of page 37- EXAMPLES. (1) A paint P given by the coordinates x= + Bft., y — + 4:ft., z = has the component velocities v, = 40, Vi = 50, Va = 60 ft. per sec., making the angles with X, Y and Z, cxi = 60°, /i, = 150% ri = 90°; a, = 120% fJ, = 30°, ^^ =90^ a, = 120 \ /J3 = 150°, r^ = 90°. Find the resultant velocity and the resultant moment about the origin. Ans. The component velocities are in one plane and «a; = + 20 - 25 - 30 = - 35 ft. per sec; % = - 34.64 + 43.3 - 51.96 = - 43.3 ft. per sec. "The resultant ®r = V'»x^ + f>y^ — 55.67 ft. per sec, making with the hori- zontal the angle cos a — ~ = pir-^x or a = 128° 57' 17", and with the vertical ^ th- 55.67 angle cos 6 = -^ = '^i^f, or 6 = 141° 2' 43". ^ Vr 55.67 The moment of the resultant velocity Vr with reference to is Mz — — 130 -f 140 = 4-10 sq. ft. per sec. The direction of motion of the radius vector in the plane XT is therefore counter-clockwise. The lever-arm p = ^^-^ — about 0. 18 ft. The equation of the line of direction of the resultant velocity isy = 1.237aj -(-0.286 ft. The intercepts on the axes are y' — +0.286 ft., of = — 0.231 ft. (2) A point P given by co-ordinates x = + 3 ft., y = + 4: ft., z = + 5 ft. has the component velocities v, = 40, Vi = 50, Vs = 60 ft. per sec, making the angles with X, Y, Z, »: = 60°, /^i = 100°, yx obtuse ; <u-2 = 100°, fii = 60°, r-i acute ; n-3 = 120°, 03 = 100°, r^ acute. Find the resultant velocity and the resultant moment about the origin. Ans. We find the angles y (page 12) by the equation cos'^ y =z — cos (or -|- (i) cos (a — ft). Hence y, = 148° 2' 31".7, y^ = 31° 57' 28".3, y^ = 31° 57' 28".3. Bj; = -f 20 - 8.6824 - 30 = - 18.6824 ft. per sec. ty —- 6.946 + 25 - 10.419 = -t- 7.635 ft. per sec Vz =- 33.937 + 42.421 + 50.907 .= + 59.391 ft. per sec 68 KINEMATICS— GENERAL PBIN0IPLE8. [CHAP. VI. The resultant velocity is «r = i/'Vx' + «j/* + Wz" = 63.73 ft. per sec, making with the axes of X, Y, Z angles given by - 18.6834 63.73 or a = 107° 19' 36"; COS 6 = ^^^2^, or & = 83° 0' 33" ; cosc = t|:||^, or c= 18" 46' 42". The moments in the co-ordinate planes are Jf^ =: + 32.905 + 74.7396 = + 97.6346 sq. ft. per sec. Mx= -]- 337.564 - 38.175 = + 199.389 " " " " My= - 93.412 - 178.173 = - 371.585 " " " " The resultant moment is Mr = V^'"' + J^y' + Mz = 350.78 sq. ft. per sec. The line representative makes with the axes of X, Y, Z angles given by Jfx 4- 199.389 COS. =^ = ^^11:^^. ore = 140-44' 8"; Mr 350.78 eos/=^ = + -'-^:^. or/= 73» 50' 31". •' Mr 350.78 ' '' Looking along this line towards 0, the motion of the radius vector is counter-clockwise. The equations of the projections of the direction of the resultant velocity Vr upon the co-ordinate planes are : on plane XY, y - - 0.408« + 5.326 ft.; on plane FZ. s = + 7.778jr - 36 115 ft.; on plane ZX, x= - 0.314z 4- 4.573 ft. The point in which the direction of the resultant velocity pierces the plane Xris given by «' = -f 4.573 ft., y' - -\- 3.357 ft. The point in which the direction of the resultant velocity pierces the plane YZ is given by y' -= + 5.336 ft., s' = -|- 14.56 ft. The point in which the direction of the resultant velocity pierces the plane ZX is given by z' = - 36.115 ft., a;' = + 13.788 ft. (.3) A point given hy the co-ordinates x=: + 3 ft., y = + 4 ft., z = 0, has the component accelerations fi = 40, /i = 50, fa = 60ft.-per- sec. per sec., making the angles with X, Y and Z, ^i = 60°, /?i obtuse, Xi — 90° ; /32 = 30°, 0-2 obtuse, y-i = 90° ; a^ = 120", A obtuse, Xi = 90°. jPmd the resultant acceleration and the resultant moment about the origin. Ans. (page 64). The component accelerations are in one plane and /a;= ~f 20 — 25 — 30 = — 35 ft.-per-sec per sec. fy= - 34.64 + 43.3 - 51.96 = - 43.3 ft.-per sec. per sec. The resultant fr= Vf^* +/"" = 55.67 ft.-per-sec. per sec, making with CHAP. VI.] ACCELERATION — EXAMPLES. 69 the liorizontal the angle cos a — -^ — ^, ^^ , or a ~ 128° 57' 17", and with / 50.67 the vertical the angle cos 6 = ^ = ^, ^" , or i* = 141° 2' 43'. / 00.67 The moment of the resultant acceieraiion with reference to is Mz = — 130 -j- 140 = + 10 square feet-per-sec. per sec. The direction of mo- tion of the radius vector in the plane XT is therefore counter-clockwise. The lever-arm n of the resultant is p = ;rir-;r;z^ = about 0.18 ft. ^ ^ o5.67 The equation of the line of direction of the resultant acceleration is y = 1.287a; -+- 0.286 ft. The intercepts on the axes are y' = + 0.286 ft., x' = — 0.231 ft. (4) A point given by the co-ordinates x = +3ft., y=+4ft.y = + 5 ft. has the component accelerations fi = 40, /, = 50, /a = 60 p;.-per-sec. per sec., making the angles with X, F, Z, a\ = 60°, fii = 100°, Xi obtuse; a^ = 100°, fi^ = 60°, y^ acute; az = 120°, fis = 100°, xs acute. Find the resultant acceleration and the result- ant moment about the origin. Ans. (page 65). We find the angles y (page 12) by cos*^ = — cos (or -)- (i) cos {a — /3). Hence y^ = 148° 2' 31 ".7, y^ = 31° 57' 28".3, r» = 31° 57' 28".3. /a, = + 20 - 8.6824 - 30 = - 18 6824 ft.-per-sec. per sec; /„=- 6.946 + 25- 10.419 =+ 7.635 " " " " " ; /2 = - 83. 937 + 42. 421 + 50. 907 = + 59. 391 " " " " « . The resultant acceleration is fr = Vfx^ +fv +f'' = 62.73 ft.-per-sec. per sec., making with the axes of X, Zand Wangles given by - 18.6824 -.nr^o .„, ofl,. cos a = — , or a = 107 19 36' ; cos6=i|;^. or 6= 83° 0' 33"; cos c = i^^. or e = 18° 46' 42". The moments in the co-ordinate planes are Mz = -{- 22.905+74.7296 = + 97.6346 sq. ft.-per-sec. per sec. Jfx = + 237.564- 38.175 = + 199.389 " " " " " " My=- 93.412 - 178.173 =- 271.585 " " " " ' " The resultant moment is Mr = V ^x" -\- Mv"* -{- Mz^ = 350.78 sq. ft.-per-sec. per sec. The line representative makes with the axes of X, 7, Z angles ^ven by , Mx +199.889 - K^oni,or,n cosd = ^= ^Q^e . or d= 55° 21' -37"; My -271.585 ..noAA> a" «°^^ = :# = -407:6-''''^ = ^^^ ^ ^ ' Mz _ +97.6346 cos/== -^ - 407 g . or /_ 73 50 21 . 70 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VI. Looking along this line towards 0, the motion of the radius vector is coun- ter-clockwise. The equations of the projection of the direction of the resultant acceleration / upon the co-ordinate planes are : on plane XT, y = - 0.408a! -f 5.226 ft.; on plane YZ. z = + 7 778^^ - 26.115 ft.; on plane ZX, x=- 0.314z -f 4.572 ft. The point in which the direction of the resultant acceleration pierces the plane XFis given hy x = -]- 4.572 ft., y' = +3.357 ft. The point in which the direction of the resultant acceleration pierces the plane YZis given hyj/=-\- 5.226 ft., g' = - 14.56 ft. The point in which the direction of the resultant acceleration pierces the plane ZX is given by 2' = — 26.115 ft., a;' = + 12.788 ft. . (5) A point given by the co-ordinates x = + 3 ft., y= + 4 ft., 2 = has the component displacements d\ = 40 ft., d^ = 50 ft., d3 = QOft., m,aJcing the angles with X, Y and Z, a, == 60°, /3, obtuse, yi = 90''; /ia = 30°, a^ obtuse, r^ = 90; as = 120°, /ia obtuse, y^ = 90°. Find the resultant displacement and moment about the origin. (6) A point given by the co-ordinates x = + B ft., y = + 4 ft., z — +5ft. has the component displacements di = 4S)ft., d^ = 50/^., da — GO ft., making the angles with X, Y and Z, ai = 60°, fii = 100°, yi obtuse; ^a = 100°, /Ja = 60°, r-i acute; «3 = 120°, ^3 = 100°, rs acute. Find the residtant displacement and moment about the origin. UA-<ir//y— -^-O'^^ < •c AT ' /\^CO . CHAPTEE VII. ANGULAR REVOLUTION OF A POINT. ANGULAR SPEED. KATE OF CHANGE OF ANGULAK SPEED. EQUATIONS OF MOTION OF A POINT UNDER DIFFERENT RATES OF CHANGE OP ANGULAR SPEED. ANGULAR SPEED IN TERMS OF LINEAR VELOCITY, RATE OF CHANGE OF ANGULAR SPEED IN TERMS OF LINEAR. MOMENT OF LINEAR VELOCITY IN TERMS OF ANGULAR SPEED. MOMENT OP TANGENTIAL ACCELERATION IN TERMS OF RATE OF CHANGE OP ANGULAR SPEED. NORMAL ACCELERATION IN TERMS OF ANGULAR SPEED. MOTION IN A CIRCLK. Angular Revolution of a Point about a Given Point.— When a point moves in any path whatever from the initial position Pi to the final position P^ in any given time, we have called the distance PiPa the linear displacement (page 34). If we choose any point in space O as a pole and draw the radius vector OPi to the initial and OP-i to the final position, we call the angle P, OP-i = the angular revolution of the point P about O. Since the angle 9 is measured in radians, it is independent of the length of the radius vec- tor, or the distance of P, and Pi from O (page 5). It is also independent of the position of the plane of revolution Pi OPi in space, or of the direction in space of the angular revolu- tion. It has, however, magnitude and sign (+) or (— ), according as the radius vector moves in this plane in one direction or the other. Angular revolution has then magnitude and sign, but not direc- tion. It is therefore a scalar quantity like distance described by a point, and cannot be represented by a straight line. The student must not confound angular revolution with "angular displace- ment," which, as we shall see hereafter (page 170), has like linear displacement, direction as well as sign and magnitude, and is therefore a vector quantity which can be represented by a straight line. Angular Revolution of a Point about a Given Axis.— The angular revolution in any given time of a moving point about a given line or axis is the angle between perpendiculars from the initial and final positions of the point to the axis. Thus let OA be a given axis, Pi and Pn the initial and final positions of the mov- ing point, and PxB, PaC perpendiculars to OA. Then the angle between PiB and P2C is the angular revolution about OA, whatever the path between P, and P^. This angle is the same as the angle pCPi, if we complete the rectangle CPi. As the straight line pPi is thus the pro- 71 72 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VII, jection of the line PiPi on the plane PiCp, we see that the angular revolution about the axis is the angular revolution of the projection p of Pi about the point C. Mean Angular Speed of a Point about a Given Point or Axis.— The angular revolution per unit of time is the mean angular speed of a- point about a given point or axis. Like angular revolution it has then magnitude and sign accord- ing to direction of motion in the plane of revolution, but is inde- pendent of the position in space of that plane. It is therefore a scalar quantity like linear speed (page 16). When the mean angular speed varies with the time it is variable. When it has the same magnitude no matter what the interval of time it is uniform. A point moving with uniform angular speed evi- dently describes equal angles in equal times. Instantaneous Angular Speed of a Point about a Given Point or Axis. —The limiting value of the mean angular speed when the interval of time is indefinitely small is the instantaneous angular speed. If the instantaneous angular speed is variable, the mean angular speed has different values for equal intervals of time. The term angular speed always signifies instantaneous angular speed unless otherwise specified. Angular speed like mean angular speed is therefore a scalar quantity, having magnitude and sign according to the direction of motion in the plane of revolution, but independent of the position of this plane in space. The student must not confound angular speed with "angular velocity," which, as we shall see hereafter (page 174), has direction as well as sign and magnitude and is therefore a vector quantity. Numeric Equations of Angular Speed.— The unit of angular speed is evidently one radian per second. We denote the magnitude of the angular speed thus measured by the letter go. If then &, is the angle measured in the plane of revolution from any fixed line to the initial position of the radius vector and to the final position of the radius vector, we have for the mean angular speed t When the interval of time is indefinitely small, we have in/the Calculus notation, for the instantaneous angular speed, dO •'=di (2> Sign of Angular Speed. — These equations are precisely the same as equations (1) and (2), page 16, simply substituting 6 for s. The sign follows the same rule. Thus when the angle is increasing the value of o) is positive (4-), and when decreasing it is negative (— ); Equation (1) is thus^ general if we take angles in any one direc- tion in the plane of revolution measured from a fixed line in that plane as positive, and in the opposite direction as negative. Angular speed, then, whether uniform or variable, mean or in- stantaneous, is independent of direction in space. It is entirely comparable to linear speed (page 15). Change of Angular Speed. — When the angular speed of a pomt varies, the difference between the final and initial instantaneous speeds for any interval of time is the integral change of angular speedy CHAP. VII.] RATE OF CHANGE OF ANGULAR SPEED. 75 Mean Rate of Change of Angular Speed. — The integral change of angular speed per unit of time is the mean rate of change of angu- lar speed. When the mean rate of change varies with the time it is vari- able. When it has the same magnitude no matter what the inter- val of time it is uniform. Instantaneous Rate of Change of Angular Speed.— The limiting value of the mean rate of change of angular speed when the inter- val of time is indefinitely small is the instantaneous rate of change of angular speed. Rate of change of angular speed should always be understood as meaning instantaneous rate of change unless otherwise specified. Rate of change of angular speed may be zero, uniform or vari- able. When it is zero the angular speed is uniform and the same as the mean speed for any interval of time. When it is uniform the rate of change of angular speed is the same as the mean rate of change for any interval of time. When it is variable the mean rate of change has different values for equal intervals of time. Numeric Equations of Rate of Change of Angular Speed. — The unit of rate of change of angular speed is then one radian-per-sec. per sec. We denote its magnitude thus measured by the letter a. If then 00 1 is the initial and co the final instantaneous angular speed, we have for the mean rate of change of angular speed GO — Gi3i " = -7-- <»> anmf or the instantaneous rate of change of angular speed "=d^ = d^ ^^> Cf Sign of Rate of Change of Angular Speed. — These equations are precisely the same as equations (1) and (2), page 25, simply substi- tuting 00 for V and 6 for s and a for a. The value then of a is posi- tive ( + ) when the angular speed increases and negative (— ) when it decreases during the time. ..i-. • It is tevident that this sign has no reference to position in space. Rate of change of angular speed is then a scalar quantity. The student must not confound rate of change of angular speed with ,y angular acceleration," which, as we shall see hereafter (page 175), has direc- tion as well as sign and magnitude and is therefore a vector quantity. Equations of Motion of a Point under Different Rates of Change of Angular Speed. — We have equations preciseljr similar to those for linear speed, page 27. We have only to substitute a? for v, a for a, 5 for s. (a) Rate of Change of Angular Speed Zero. — In this case if Qi is the initial angle measured in the plane of revolution of the radius vector from a fixed line in that plane, and 6 the final angle, we have 6 — Oi . ^ « /-4 ^ GO = , or cot = 6 — Qi (1) Revolution in any one direction in the plane being taken as posi- tive, in the other direction it is negative. Then if &? comes out (-f-) it denotes revolution in the assumed positive direction; if (— ), in the opposite direction. If t comes out (+) it denotes time after, if (— ) time before, the start. 74 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VII. (b) Rate of Change of Angular Speed Uniform. — When the rate of change of angular speed is uniform, the instantaneous rate of change of angular speed at any instant is equal to the mean rate of ■change for any interval of time. If fi9i and GO are the initial and final instantaneous angular speeds, we have then for the rate of change of angular speed GO — COi " = -!-- «> The value of a- is (+) when the angular speed increases and (— ) when it decreases during the time. From equation (2) we have GO = GOi + at (3) The average angular speed is 2 — = °^' + 2 ^^ The angle described in the time t is Q - 0, = .^L±^t = cod + lat\ ... (5) Inserting the value of t from (2) 'we have »-»'=-^- (« Hence oj" = gox^ + 2a{^ — 0,) (7) In applying these formulas, a is positive (+) when the angular speed increases and negative (— ) when it decreases during the time, without regard to direction of revolution. If angles 6, Oi in one direction are taken as (+), angles in the opposite direction are (— ). Angular speeds aj, (Wi are positive (+)when motion is in the assumed positive direction, and negative (— ) when in the other direction. A positive t denotes time after the beginning of motion, and a negative t time before. [(c) Rate of Change of Angular Speed Variable] — If the rate of change of angular speed is variable, we have from (1), in Calculus notation, '^ = av (8) doo d^B ^=dt=d^' (9) and from (8), - Oi = / oodt (10) The preceding equations can be directly deduced from these as on page 28^ Angular Speed in Terms of Linear Velocity. — If a point moves A from Pi to Pi in any path in the time t, the ^ linear displacement in that time is the " ^ .Rj chord P1P2. Pj^i- '(py^ If we take O as a pole, the angular dis- placement of revolution is P1OP2 = 9. The mean linear velocity is :^ '^ chord P^P^ , . / and the mean angular speed is —. n CHAP. VII.] ANGULAR SPEED IN TERMS (HP LINEAR VELOCITY. 75 From Pi draw PiN perpendicular to OP^i. Then if angle PiP^N = 0, we have PiN= chord PiPi . sin 0. Dividing by t, we have / PiN _ chord P iPa . sin (f> t ~ t '"' But PiN = r sin 0, where r is the radius vector OPi. Hence rsin6 _ chord Pi Pa . sin t ~' i ■ If now the time is indefinitely small, -—- — ? becomes the in- I stantaneous velocity v, and cp becomes the angle APiV = e between ; the radius vector OPi and the instantaneous velocity v, and sin 9 ; becomes S, and —becomes the instantaneous angular speed oo. Hence j usine „. rcj = V sm e, or go = . (1) r In general, then, whatever the path or wherever the pole. The magnitude of the angular speed at any point is equal to the magnitude of the component of the linear velocity at that point per- pendicular to the radius vector, divided by the magnitude of the radius vector. If the pole O is taken at the centre of curvature, so that OPi is equal to the radius of curvature p, then e = 90° and we have poo = v v or o) = — . P Rate of Change of Angular Speed in Terms of Tangential Linear Acceleration. — If ft = a is the magnitude of the linear tangential acceleration or rate of change of speed at any point, then we can prove, precisely as in the preceding Article, that a = fi^B:I (2) r where a is the magnitude of the rate of change of angular speed. Hence the magnitude of the rate of change of angular speed at any point is equal to the magnitude of the component of tJie linear tangential acceleration at that point perpendicular to the radius vector, divided hy the magnitude of the radius vector. If the pole O is taken at the centre of curvature, e = 90° and we have pa — ft or a = — . Moment of Linear Velocity in Terms of Angular Speed.— We can resolve the linear velocity v at the point P into a component v cos e along the radius vector and a component ^ v sin e perpendicular to the radius vector. Xg The moment of the first relative to the pole ^ -^ Z^ — Uv is zero. Since the moment of r is equal to ; / \"~"~"^ the algebraic sum of the moments of its com- ^j Z \ ^ ponents (page 62), if we take moments about ! / the pole, we have ]/^ o t}p = V sm € . r. 76 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VII. But we have just seen that v sin e = roo. Hence . vp = r'^oo (3) That is, the magnitude of the moment of the linear velocity at any /-, point relative to the pole is equal to the magnitude of the angular / f ]/t/* speed at that point, multiplied by the square of the magnitude of the k radius vector. Since v sin e is the normal component of v, v sin e . r is twice the areal velocity of the radius vector (page 61). Moment of Linear Tangential Acceleration in Terms of Rate of Change of Angular Speed. — We can resolve the tangential accelera- tion /( into components along and perpendicular to the radius vector and thus obtain, precisely as in the preceding Article, fP=ftPt='^''' (4) Hence the magnitude of the moment of the linear tangential ac- celeration at^ any point relative to the pole is equal to the magnitude of the rate of change of angular speed at that point, multiplied by the square of the magnitude of the radius vector. Since ft sin e is the normal component of ft , ft sin e .r = r''a is twice the areal acceleration of the radius vector (page 61). Normal Linear Acceleration in Terms of Angular Speed. — We have seen (page 53) that when a point moves in any path, the v'^ magnitude of the normal acceleration f^ is given by f^ = — , where p is the radius of curvature. If we take the pole at the centre of curvature, then, we have v — pa, and hence /n = 77 = P*"' = ^'^ (5) The magnitude of the normal linear acceleration at any point is equal to the magnitude of the radius of curvature at that point, multiplied by the magnitude of the square of the angular speed, or to the velocity voa in the hodograph (page 52). Since for any path roo = v sin e, we have Hence, in general, roo sin e . . V' r'ftj' f =VGO = pCO^ = — = P p sm" € where r is any radius vector when the pole is not at the centre of curvature, and e is the angle of v with this radius vector. Motion in a Circle. — For a point moving in a circle we have e = 90° and r = p. Hence from (1), page 75, we have V v = roo, or 00= — (1> r If r is unity, we have the numeric equation go = v, that is, the number of units of angular speed is equal to the number of units of linear speed at distance unity. From (2), page 75, we have ff = ra, or a= — (2) -CHAP. VII.] EXAMPLES — ANGULAR SPEED. If r is unity, we have the numeric equation a =ft, that is, the number of units of rate of change of angular speed is equal to the number of units of linear tangential acceleration at distance unity. From (5), page 76, we have in any case fn = 'f^^^ (3) or the normal linear acceleration is equxil to the velocity in the hodograph (page 53). Inserting the value of v from (1), fn = roo'=~^ or c.^=-^ (4) If r is unity, we have the numeric equation fn = v'^ = oo^, that is, the number of units of the normal linear acceleration is equal to the square of the number of units of linear velocity at distance unity ; or the square of the number of units of angular speed is equal to the number of units of the normal linear acceleration at distance unity. We have also for the total resultant linear acceleration / = Vfi' +fn\ (5) Since the component fn passes through the centre, the moment of / relative to the centre is equal to the moment of /^. Hence w — r^co, (6) fp=ftr = r'a (7) give the moments of v and / with respect to the centre. If the point starts from rest and acquires the velocity v in the V time t, under constant tangential acceleration, we have /^ = — , GO a = —. t Graphic Representation of Rate of Change of Angular Speed. — We can represent intervals of time by distances laid oflE horizontally and the corresponding angular speeds by distances laid off vertically and thus obtain the same diagrams as for linear speed given on page 29. EXAMPLES. (1) The angular speed of a point moving in a plane about some assumed point changes from 50 to 30 radians per sec. in passing through 80 radians. Find the constant rate of change of angular speed and the time of motion. ca' — aji' ,. mi . . T Ans. a = -—- ^— - = — 10 radians-per-sec. per sec. The minus sign de- 2(e — 6i) f r notes decreasing speed, t = = 3 sec. (2) Draic a figure representing the motion in the preceding example, and deduxie the results directly from it. Ans. Average speed = — '^ — = 40 radians per sec. Hence 40i = 80 or < = 3 seconds. Also '•'r* ^ 30-50 ,_ ,. '^-*^ a. = = — 10 radians-per-sec. per sec. 78 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VII» (3) A point mooing in a plane has an initial speed of 60 radians per sec. about an assumed point and a rate of change of speed of + 40 radians-per-sec. per sec. Find the speed after 8 sec; the time required to describe 300 radians ; the change of speed ivhile describ- ing that angle; the final speed. Ans. See Example (9), page '31. (4) If the motion in the last example is retarded, find (a) the angular revolution from the start to the turning-point ; (b) the angle described from, the start after 10 sec. ; the speed acquired and the angle between the final and initial positions ; (c) the angle described during the time in which the speed changes to — 90 radians per sec, atid this time ; id) the time required by the moving point to return to the initial position. Ans. See Example 10, page 32. (5) A point moving in a plane describes about a fixed point angles of 120 radians, 228 radians and 336 radians in successive tenths of a second. Show that this is consistent with uniform rate of change of angular speed, and find this rate. Ans. a — 10800 radians-per-sec. per sec. (6) Two points A and B move in the circumference of a circle with uniform angular speeds go and go'. The angle between them at the start is a. Find the time of the nth meeting, the angles described by A and B, and the interval of time between two successive meet- ings. Ans. See Example (21), page 21. Time of the wth meeting, tn = ' ; • CO ± GO Angle described by .4 is 6 = cotn. " " S is T a. Interval of time between two successive conjunctions is \A' ' _^-/~r CO ± GO where we take the (+) or (— ) sign for a according as B is in front of or behind A at start, and {-\-) or (— ) sign for go' according as the points move in opposite or the same directions. (7) What is the angular speed of a fly-wheel 5 ft. in diameter ivhich makes 30 revolutions per minute, and what is the linear ve- locity of a point on its circumference f Also find its linear normal acceleration and the moment of its velocity with reference to the centre. Ans. Tt radians per sec; 2.57r ft. per sec, tangent tocirc; 2.5;r'^ ft.-ptr- sec. per sec; 6.25;r sq. ft. per sec. (8) Find the linear and angular speed of a point on Hie earth's equutor, taking radius 4000 miles ; also the linear normal accelera- tion. Tt Ans. 1535.9 ft. per sec; -— radians, or 15° per hour; 0.112 ft.-per-sec per sec. "%. (9j The angular speed of a wheel is —Tt radians per sec. Find the linear speed of points at a distance of 2ft., 4 ft. and 10 ft. from the centre, also the linear normal acceleration. ■^k. i, CHAP. VII.] EXAMPLES — ANGULAR SPEED. 79 g Ans. ^ ic, Zic, 7. 5;r ft. per sec. a 9 9 45 ^-nr", -j-ffS -^-tt'^ ft.-pei-sec. per sec. o 4 o (10) ijf the linear speed of a point at the equator is v, find the speed linear and angular at any latitude A. n Ans. « cos A ; -— radians per liour, or 15° per hour. (11) A point moves with uniform velocity v. Find at any instant its angular speed about a fixed point whose distance from the path is a. Ans. -i- radians per sec, where r is the radius vector. Uniform velocitr r" means uniform speed in a straight line (page 43). Hence the angular speed of a point moving with uniform speed in a straight line is inversely proportional to the square of the distance of the point from a fixed point not in the line. (12) The speed of the periphery of a mill-wheel 12 feet in diame- ter is 6 feet per sec. How many revolutions does the wheel m^xke per sec. f Ans. jr — revolutions. (13) The time i^ between 5 and 6 d'cloch, and the hour and min- ute hands are together. What is the time f Ans. oh. 27 m. 16 sec. (see Example (6) ). (14) Express in degrees and radians the angle made by the hands of a clock af 3.35 o'' clock. Ans. 102.5 deg. , 1.79 radians. (15) Find the multiplier for changing revolutions per minute into radians per second. Ans. 0.10472 rad. per sec. = 1 rev. per min. (16) The minute and second hands point in the same direction at 12 o'clock. When do they next point in the same direction f Ans. 1 min. Ig'^ sec. after twelve. (See Example (6)). (17; Two clocks are together at XII. When the first comes to /, it has lost a second ; when the second comes to I, it has gained a second. How far are they apart in 12 hours f Aus. 24 sees. ik (18) Two men start together to walk around a circular course, one taking 75 minutes to the round, the other 90. When will they be to- gether again at the starting-point f Ans. 7.5 hours. (See Example (6)). \ (19) Tlie hour-hand of a watch is f of an inch long, the minute- '^' hand f of an inch, and the second-hand i of an inch. Compare the lineal speeds of their points and the angular speeds. Ans. 5 : 112 : 2800; 1 : 12 : 720. (20) Deduce the equivalent of longitude for one minute of time and for one second of time. Ans. 15' to 1 min., 15" to 1 sec. so KINEMATICS— GEKERAL PRINCIPLES. [CHAP. VII. (21) TTie diameter of the earth is nearly 8000 miles. Required the circumference at the equator and the linear speed at latitude 60°. Ans. 25000 miles; 521 miles per hour. ' ' (22) The wheel of a bicycle is 52 inches in diameter and performs 5040 revolutions in a journey of 65 minutes. Find the speed in miles per hour ; the angular speed of any point about the axle ; the areal velocity of a spoke ; the relative velocity of the highest point with respect to the centre. Ans. 12 miles per hour; 8.12 radians per sec; 19.06 sq. ft. per sec; 12 miles per hour. (23) In going 120 yards the front wheel of a carriage maJces six revolutions more than the hind wheel. If each circumference were a yard longer., it would make only 4 revolutions more. Find the cir- cumference of each wheel. Ans. 4 yards and 5 yards. (24) If the velocity of a point is resolved into several components in one plane, show that its angular speed about any fixed point in the plane is the sum of the angular speeds due to the several compo- nents. (25) A point moves with uniform speed v in a circle of radius r. Show that its angular speed about any point in the circumfer- . V ence ts — . 2r (26) Show that the angular speed of the earth about the sun is proportional to the apparent area of the sun's disk. [The radius vector from the sun to the earth sweeps over equal areas in equal times.] W (^ (27) A point P moves in a parabola with constant angular speed \i ii\ about the focus S. Show that its linear speed is proportional to tl-i. ySP- ^.L V (28) A point starting from rest moves in a circle with a constant rate of change of angular speed of 2 radians -per-sec. per sec. Find 'j»^ the angular speed at the end of 20 sec. and the angular displacement ^ of revolution ; also the linear speed and distance described and Si -^ the number of revolutions ; also the linear tangential acceleration r and the normal linear acceleration at the end of 20 sec. Ans. 40 rad. per sec; 400 radians; 40r ft. per sec; 400rft. ; - — revolu- tions; 2r ft.-per-sec. per sec. tangential acceleration; 1600r ft.-per-sec per sec. normal acceleration. (29) A point moving with uniform rate of change of angidar speed in a circle is found to revolve at the rate of 8i revolutions in the eighth second after starting and 7i revolutions in the thirteenth second after starting. Find its initial angular speed and its uni- form rate of change of angular speed ; also the initial linear speed and rate of change of speed; also the initial normal acceleration. Ans. 20.27r radians per sec; — 0.47r radians-per-sec per sec; 20.2;rr ft. p^sec; — 0.47rr ft.-per-sec. per sec; 408.047rV ft.-per-sec per sec. ly (30) A point starts from rest and moves in a circle with a uni- form rate of change of angular speed of 18 radians-per-sec. per sec. Find the time in which it makes the first, second and third revolu- tions. Ans. —^ — ' -z ' t: beoB. 3 3.3, m ^ / CHAPTEE VIII. DIFFERENTIAL EQUATIONS OF MOTION OF A POINT.* Free Motion of a Point — Rectangular Co-ordinates.— Let a mov- ing point have a position at any instant given by the co-ordinates x, y and z, and let the distance described in the interval of time dt be ds, and let the direction of motion make the angles or, yS, y with X, F, Z. Then we have dx „ dy dz cos a = —-, cos p = -^ , cos y = t-. ds ds ds The magnitude of the velocity is V = ds (1) (3) and its components in the direction of the axes are ds dx Vx='0 cos a = -- cos nr = — -; dt dt' „ ds ^ dy ds dz (3) > 1 r ■ » '^ >Vx "6 ■ ■■ .Vz X y / U " ■»« = t> COS V = -7- cos r = ,^ dt dt We have Vx positive towards the right, negative towards the left; % posi- tive upwards and negative downwards ; Vz positive in the direction OZ, negative in the opposite direction (page 44). Squaring equations (3) and adding, since cos" a + cos' yS -1- cos' ;k = 1, we have . = ...v.v = (§)-=(|)V(f)%(D-,..(.) or dt ^ \dt) + l^J -^ (l)° dy (5) Let the acceleration be / and its components in the direction of the axes X, F, ZhQfx,fy,fz, then we have = ^. f.= d'z (6) * This Chapter must be omitted by those not familiar with the Calculus. 81 82 KINEMATICS— GENERAL PRINCIPLES. [CHAP. VIII. The acceleration / is then We have fx positive towards the right, negative towards the left; fy positive upwards, negative downwards; fz positive in the direction OZy negative in the opposite direction. The tangential acceleration ft = a is, the rate of change of speed, or „ dv (Ps ^„. ^' = '' = M=dt^ (^> Differentiating (4) and substituting (6), we have vdv =fxdx + fydy + fzdz. Hence r ^ . . . (9) tr* = 2 / ifxdx + fydy + fzdz) + Const. cLs d^s Dividing by ds, since v = -^, and dv = -— - , we have from (8) „ dv cPs ^clin „ dy dz dx d'x dy dj^y dz d?z ,^„. d» df ds dt^ da dC ^ ^ The normal acceleration /„, as we have seen (page 76), is /„ = — , where P p is the radius of curvature at the point. We have from analytical- Geometry p" \d»^ j \ds* J \as / Hence and the acceleration/ is f=Vfi'+fn'. (11&) If we denote by S the angle which the acceleration / makes with the radius of curvature p, and by e the angle which it makes with the tangent to the curve, we have S + e,= 90° and tan e = ^ = -^'- (12) '''m The moment of the velocity about the origin is the sum of the mo- ments of the components. The moments in planes parallel to XT", YZ, ZXaxQ : about s, Mz= —■ X — — -y; dt dt ., ,. <7s dy " X, Mx= — ■ V — — • z: "^ dt " dt ' dx dz dt '^~ di My = -rr ■ e — —. ■ X. (13a). CHAP. VIII.] DIFFERENTIAL EQUATIONS OF MOTION OF A POINT. 83 The moment about the origin of the resultant velocity t? if j? is its lever-arm is then vp = M= i/Mx"" + My" + mF. (186) The line representative of this moment makes the angles d, e, f with the axes of JT, F, Z given by J Mx My ^ Mz cosd = — , cose = -^, cos/ =-- (14) Looking along this line representative towards the origin, the direction of rotation is always counter-clockwise. In the same way the moment of the acceleration about the origin is the sum of the moments of the components. We have then precisely the d?x d?v d?z same equations as (13), (14), only we put — -, — f-, ■— in place of djf 0/L aXf (lOR O/U o& -r-y -^, 3- in order to find the moments in the co-ordinate planes. at at at Application of the Preceding Formulas.— Equations (3) and (6) are the general equations by which the motion of a point is determined. Applications of the use of the equations just deduced will be given here- after. We can only indicate here the general application. If ^ = 0, we have motion of a point in a plane only. The correspond- ing equations are at once obtained by making z and dz zero wherever they occur in the general equations. If we also make /3 = and ^ = 0, we have motion along the axis of x only. Hence taking x = s, we have from (3) v=-; from (6), / = /, = « = —; which are equations (8), (9) of page 51. If the velocity v in any case is given, it can be resolved by (3) into its components Vx, Vy , %. Then by differentiating as indicated by (6) the components /x,fy, fz of the acceleration / can be found, and the accelera- tion / can then be found by (7). If the component accelerations are given, we find by integration the component velocities and then the resultant velocity. If the path is required, each of equations (6) must be integrated twice, thus introducing two constants of integration for each. The constants of the first integration will depend on the initial velocity, those of the second on the initial position. We thus obtain equations involving x, y, z and t, and by eliminating t we obtain an equation between x and y, or y and ^, or z and a;, that is, the equation of the projection of the path ou the co-ordinate planes. Differential Polar Equations for Motion of a Point in a Plane. — Let X and y be the rectangular co-ordinates, and ^_^ r and 9 the polar co-ordinates, of a point P in a / ""^-.^ plane. Then P/ " ic = r cos 6, y = r sin 9 (15) Differentiating and dividing by dt, we have dx dr . • r,dQ • ,^-, «?«=-- = —-cose — rsme —: (16) dt dt dt' dy dr . . .dQ .^„. % = -^ = -— smfl -hrcose— - (17) ^ dt dt dt ^ ' 84 KINEMATICS— GENERAL PBINCIPLES. [CHAP. VIII. Squaring and adding, since sin' 6 + cos" 6 = 1, we have for the magni- tude of the velocity -(§)■= (0-'(f)' (-> If r is constant, the path is a circle. In this case — - is zero and v = at r——= roo, where oo is the angular speed (page 76). The velocity along the radius vector is -TT = -TT cos e + -/- sin 9 (19) dt dt dt The velocity perpendicular to the radius vector is i"-^ = -jf cos fl — sin 9 (20) dt dt dt fj or fl^'}i Since by (6) -— and -~ are the horizontal and vertical components of the acceleration, we have, by differentiating (16) and (17), ^^=w=\dt^-i-dt) \''''-['di-dt'-'dPi''''''^ ^'i> •^^=d^=\dF-idFi r^'-'rczF^+'^J''''' ^''^ The acceleration along the radius vector is then fx cos ^ +fy sin 9, or ''''^d^''''=d^-idi) ^^^> If r is constant, the path is a circle. In this case -— is zero, and the acceleration along the radius vector is /n = — rco^, where go is the angular speed (page 76). The (— ) sign denotes direction towards the centre (page 50). The acceleration ^erpewc^icwZar to the radius vector is /j,cos9 —/a; sin 9, or ^cos9 — TV sm9 = 2— — + r-— - (24) dt^ dt'' dt dt dt"" ^ ' If r is constant, the path is a circle, and — is zero, and the acceleration perpendicular to the radius vector is ft = ra, where a is the rate of change of angular speed (page 76). Equation (24) may be written ^dtdt'-'-df = i:diT'di) '^'^ From equation (13a) we have, by inserting the values of x, y and ^-, dt d'y dt CHAP. VIII.] DIFFERENTIAL EQUATIGlJrS OF MOTION OF A POINT. 85 dy -^ from (16), (17), for the moment of the velocity with reference to the pole, if p is the lever-arm, vp = M^=r'-=r'co, (36) where oo is the angular speed (page 76). From equations (14) and (21) and (22) we have in like manner, for the moment of the acceleration, „ l^dr de <fe\ ^^=i'dfdT'-'de) <2^> We see from (25) that this may also be written d(rm fp=:A^=f^^ = r'a, (28) where ft is the tangential acceleration and a is the rate of change of angular speed (page 76). Applications of the use of these formulas will be made hereafter. General Polar Equations of Motion of a Point in a Plane — Ac- celeration Central. — When the acceleration is always directed to or from a fixed point it is called central acceleration, and the fixed point is called the centre of acceleration. Let this fixed point be the pole. Then, since the direction of the acceleration always passes through the pole, its moment with reference to the pole is zero, and we have from (28) i"^)='>' "'"§=' w where c is a constant of integration. dQ Now -5— = oa = angular speed, and from page 75 we have at roo = v sin e, wiere e is the angle which the velocity p at any point makes with the radius vector. — r" Therefore '^ r" — = r'oo = rv sm e — e. 5-7^ — ' — P dt f^\° I From page 76 we see that r^ta is the mo- \ g | ment of the velocity and is equal to twice the ^X" areal velocity of the radius vector. \T Hence in central acceleration, the moment of the velocity about the pole is constant, the area described by the radius vector in a unit of time is constant, and tJie radius vector tJierefore de- scribes equal areas in equal times. The constant c is twice the area described by the radius vector in a unit of time (page 61). If p is the perpendicular let fall from the pole upon the direction of the velocity, we have vp = r" — = r'fij = rv&m € = e (30) ^ dt From (30) we have do c , c . — —00=:.— and v = — (31) dt r P 86 KINEMATICS— GENERAL PRINCIPLES. [CHAP, VIII. Hence for central acceleration the angular speed at any point of the path is inversely as the square of the radius vector, and the linear velocity at that point is inversely as the perpendicular distance from the centre of acceleration to the tangent to the path at that point. If / is the central acceleration along OP, then the component of / nor- toial to v is/n=/sin e, or, since from (30) sin e = —fn =/— . But from r r v^ page 53 we have seen that fn = — , where /a is the radius of curvature. P Hence /^ = ^, or v^ = 2fxU2pA r f} 4:\ r J But 2p sin e = 2p— is the length of the chord of curvature 2FB through the pole. (See figure page 85.) From page 28 we have for a point moving from rest with uniform rate of cliauge of speed a, if = 2a(s — Si). Therefore, for central accelera- tion the speed at any point of the path is equal to that acquired by a point moving from rest with constant rate of change of speed f through a space equal to one fourth the chord of curvature through the centre of acceleration. If the acceleration is central, its component perpendicular to the radius vector is zero, since the pole is the centre of acceleration, and we have from (24) 4I-I^=« <-) The component along the radius vector is equal to the acceleration itself, if the pole is the centre of acceleration, and we have from equation (23), if the acceleration is towards the centre, cPr fdQV . „„. where the (— ) sign for /denotes motion towards the centre (page 50). Equations (32) and (33) express all the conditions of central accelera- tion towards the pole, and therefore determine the motion. If the acceleration is away from the pole we have + / instead of — / (page 50). From (30) we have d^_^ df ~r*' and equation (33) becomes ^=^-/=---/- w But we have seen, page 76, that roo' is the central acceleration for a point moving in a circle of radius r with the speed roo. d?r The rate of change of length of the radius vector ^, we see from (34), is then the difference between the central acceleration /at any instant and the central acceleration at the same instant of a point moving in a circle •of radius r with the same angular velocity. This rate of change of velocity along the radius vector is called the CHAP. VIII.J DIFFERENTIAL EQUATIONS OF MOTIONOF A POINT. 87 dr paracentric acceleration. Its integral or t^ is the velocity of approach or recession along the radius vector and is called the paracentric velocity. (a) To find the speed at any point of the path. — Central acceleration. If we multiply (32) by rdB and (33) by dr and add, we have or Integrating, we obtain — dr- = ''-^J^^^^ (^«> ■where Ci is a constant of integration. If the law of variation of / is given in terms of r for any given case, we can perform the integration denoted by / fdr. From (18) we have ._d^_ dr" + y-'de' ~ dt^~ df Hence ^ = c, - 2 Cfdr (37) "We have also from (30) v = - (38) * P Since in (37) the value of v depends only upon r, we see that the speed J'or central acceleration at any two points of the path is independent of thepath^ and is the same for any points equally distant from the centre, the law of acceleration remaining the same. {b) To find the time of describing any portion of the path.— Central accelera- tion. Substituting (31) in (36), we have %^^ = '>,-iffdr. (39) Hence dr We have also from (31) t = 1 pi^dQ, (41) cje, from which r must be eliminated by means of the equation of the path and the integration performed in reference to 9. (c) To find the equation of the path. r*dO'* Substitute in (39) for dt^ its value from (31), df" = -^, and we have ,= r ";_ ,40) i^*h)='-'P''^ <^ 88 KIKEMATICS — GENEEAL PRINCIPLES. [CHAP. VIII. This equation may be simplified by putting r = - -, and it then becomes (cf) To find the law of the acceleration for any given path. — Central accelera- tion. Differentiating (43) With reference to dB, we have •^=M'^'^'') ^^^^ r*dB^ Substituting in (33) for dt"^ its value from (31), df^ = —^y we have ^=?[r-^^l' ^''> which Is the same as (44) if we put — = u. r T*dft^ From (30) we have p" = -^, and from (18) i?d^ = (f r* + r'dS'. Therefore di^ + r^dS' Substitute this in (42) and we have ^ = c, - '^i J fdr. (47) Differentiating with respect to p, /=^. . . : (48) The law of acceleration is given by (44) or (45) or (48). From (33) and (35) we have ^,g^_2_dr^ (49> r an expression which will often be found useful in reductions. Differential Equations for Constrained Motion of a Point in a Plane. — For free motion of a point in a plane we have from (6) for the horizontal and vertical components of the acceleration Under the action of these components, the point, if free to move, de- scribes some curve. But if it is constrained to move in a given curve, these components will be changed by reason of the normal acceleration N due to the given curve. If thus /is the acceleration of a free point P, fx and /j, its horizontal and vertical components, N the normal acceleration due to the given V f curve, and 6 the angle of the tangent at P u {y I with the horizontal, we have 4 .U g-=/.-Mne; .. . . (50) ^ ^=/i/ + iVcose. . . . (51) CHAP. VIII.J DIFFERENTIAL EQUATIONS OF MOTION OF A POINT. 89 If iV is zero, the motion is unconstrained and we have (6). These two equations, together with the equation of the given curve, are sufficient to determine the motion completely. In applying them, /a; is positive towards right, negative towards left, and the horizontal component of iV follows the same rule. We heiwe fy positive upwards and negative downwards, and the vertical component of JV follows the same rule. If we multiply (50) by 2dx and (51) by 2dy and add, we have, since sin G = — and cos 6 = ^-, as OS ?^^^^^^^=2(/.efc.+/,rfy). The first member of this equation is the differential of -jfi^ =^ •— = «", or is equal to 'iivdv. Hence vdv =:fxdx +fydy, ' 1 or !■• • • (^2) «' - - - - - - - »' = 3 fWxdx +fydy) + Constant. J This is precisely the same result as that obtained for free motion, equa- Hence we conclude that if there is no acceleration except that of iVdue to the curve alone, or if /x = 0, /j, = 0, the speed on the curve is constant and unaffected by the curve. If there is an acceleration besides that due to the curve, the speed will be unaffected by the curve and the same as if the point were free. If the acceleration of the free point is parallel to the axis of y, we have /x=0 and «« = 2 ffydy + Constant (53) If in this last case/y is constant, we have «« =2fyy + Constant (54) If the distance of the point from the origin y = Si when « = t?i , we have ^ ,__. «"=«,» +2/j, (2/- «.), (55) which is precisely the same as for uniform rate of change of speed for a free point, as given by eq. (7), page 56. , . , ... Regard must be had to the signs in applying these equations to any special case. Velocity and acceleration upwards are positive, downwards negative ; to the right positive, to the left negative. (a) To find the time of motion of a point on a given curve. In all cases rf< = — . Hence when the nature of the curve and the V speed at any point of it are known, the value of v may be found from (52) and substituted, and then t may be found by integration If the acceleration of the point is constant and equal to /and parallel to y, we have from (55) dt= , ^ (56) 4/®l' + 2/y(y-8l) ^0 KINEMATICS — GENERAL PRINCIPLES. [CHAP. VIII. (b) To find the normal acceleration due to the carve. If we multiply (50) by ~ and (51) by j- and subtract, we have, since sin e = -^, cose = -;-, and dx^ + dy^ = ds'', ds ds ''''ds ''^ds dsdf ds Eliminating dt by the equation « = — , we have dt dy r dx dxd-'y - dyd"" But if p is the radius of curvature of the constraining curve at the point X, y, d^ dxd^y — dyd^x ' Hence ds ds p =/a, sin e —/j, cose H , (57) The first two terms give the normal component of / for free motion. The last term is the normal acceleration due to the curve. If fx and fy are zero, the only acceleration is that due to the curve and N=~ (page 76). KINEMATICS OF A POINT. TRANSLATION. CHAPTEE I. RECTILINEAR MOTION OF TRANSLATION. FAIililNG BODY, ACCELERATION INVERSELY AS THE SQUARE OP THE DISTANCE. Translation. — We have defined translation (page 13) as motion of a rigid system, such that every straight line joining any two points remains always parallel to itself. The paths of all the points are therefore parallel at every instant and equal for any given interval of time, and the velocities of all the points at any instant are equal and parallel. If these velocities are uniform, that is, if all points move in parallel straight lines with equal speed, the translation is uniform. If these velocities change either in magnitude or direction, the translation is variable. When, then, a body has motion of translation only, the motion of the body is the same as that of any one of its points, and the study of the kinematics of a point is therefore the study of the translation of a body. Rectilinear Motion. — If the direction of the acceleration of a point does not change and always coincides with the direction of the velocity, then the velocity may change in magnitude but can- not change in direction, and we have motion in a straight line. In such case the magnitude of the velocity is the speed in a straight line, the magnitude of the acceleration is the rate of change of speed and may be either uniform or variable, and the equations of pages 28 or 51 apply. Acceleration is Proportional to Force. — Although we are now studying change of motion without reference to its cause, it will be well for the student to keep in mind the fact that no material body can change its own motion. Any change of motion is always found to be due to the action of other bodies. This action of external bodies upon the body considered to which change of motion or ac- ■celeration is due is called force. 91 92 KINEMATICS OF A POINT — TRANSLATION. [CHAP. I» The student may figure to himself such a force as the pressure or pull of an imponderable spiral spring upon the body, the axis of the spring having always the direction of the acceleration, and the spring moving with the body, so that its pressure or pull is exerted during the entire time of acceleration and is always proportional to the acceleration. If the acceleration changes in direction, the axis of this spring changes, so that it is always in the same direction as the accelera- tion. If the acceleration changes in magnitude, the pull or push of the spring changes correspondingly. If the acceleration is uniform, that is, does not change either in direction or magnitude, the axis of the spring does not change in direction and its pull or push is constant. The force of gravity upon bodies near the surface of the earth is like the action of such a spring. Its action is practically constant in intensity and direction. The student should note that the direction of the force or accel- eration is not necessarily that of the motion, except in the case of rectilinear m.otion. Thus in the case of a point moving with uniform speed in a circle, the direction of motion at any instant is tangent to the circle, but the acceleration is always directed towards the centre. Central Acceleration. — When the acceleration is thus always directed towards or away from a fixed point, it is called central ac- celeration, and the fixed point is called the centre of acceleration. If the direction of the acceleration is towards the centre, the ac- celeration is negative (page 50) and the force attractive. If away, it is positive and the force repulsive. Uniform Acceleration — Motion Rectilinear — Force Attractive. — When the direction of the uniform acceleration coincides with that of the motion, we have motion in a straight line with uniform rate of change of speed, and equations (2) to (7), page 28 or 51, apply. The most common instance of such motion is that of a body fall- ing freely near the earth's surface.* In this case the acceleration due to gravity is known to be practically constant and is always denoted by g. We have then simply to replace a or / by gr in equa- tions (2) to (7), page 28 or 51. We shall take g = 32.2 ft.-per-sec. per sec. or 981 cm.-per-sec. per sec. unless otherwise specified. Value of g. — The value of g is usually given in feet-per-sec. per sec. or in centimeters-per-sec. per sec. It has been determined by much careful experiment and found to vary with the latitude A and the height h above sea-level. * Strictly speaking there is no known instance in nature of a uniform ac- celeration (or of a force wliich does not vary in magnitude and direction). The acceleration g due to gravity (or the force of gravity) varies inversely as the square of the distance from the centre of the earth for a body outside tin; earth, and directly as the distance for a body inside, i.e., in a shaft or well. But, as we shall see, the variation due "to this cause is insensible for all ordinary distances. The decrease of ^ at a distance of a mile above the earth's surface is only about the 2000th part of its value at the surface. Also two radii of the earth are sensibly parallel when near together. It is therefore customary and practically correct to speak of ^r as a constant acceleration at any place. It should be borne in mind, however, that even then the resistance of the air very materially modifies the results for falling bodies. We can therefore only assume g as constant for fall in vacuo. CHAP. I.] RECTILINEAR TRANSLATION". 93 The general value is given by g = 32.173 - 0.0821 cos 2A. - 0.000003^, where h is the height above sea-level in feet, and g is given in feet- per-sec. per sec, or g = 980.0056 - 2.5028 cos 2A - 0.000003^, where h is the height above sea-level in centimeters and g is given in centimeters-per-sec. per sec. It will be seen that the value of g increases with the latitude, and is greatest at the poles and least at the equator. It also decreases as the height above sea-level increases. The following table gives the value of g at sea-level in a few localities : 9 9 Latitude. F. S. Units. C. S Equator New Haven Latitude 45° Paris London Greenwich Berlin Edinburgh 55 57 Pole United States ] For calculations where great accuracy is not required it is cus- tomary to take gf = 32 ft.-per-sec. per sec. or g = 981 cm.-per-sec. per sec. For the United States g = 32^ is a good average value and is therefore very often used. In exact calculations the value of gfor the place must be used. Formulas for a Body Projected Vertically Up or Down. — We have then, for a body projected vertically upwards in vacuo, simply to put —gia. place of / in equations (2) to (7), page 51. We thus obtain ' Latitude. F. S. Units. C. S. Units. 0°0' 32.091 978.10 4118 32.162 980.284 45 32.173 980.61 48 50 32.183 980.94 5140 32.182 980.889 5129 32.191 981.17 52 30 32.194 981.25 55 57 32.203 981.54 90 32.255 983.11 U9 32.162 980.26 25 32.12 979.00 v — Vi — gt; Vi — V t = +Vi' f-^ V + Vi, . 1 S — Si = g t = Vit — -i (1) (2) (3) S — Si . _ 2(s - si) _ Vi ± Vvi'' - 2flr(g - si) . ^ ^^^ ~ V + Vi g v" = Vi» - 2g(s - Si) ; (5) ^'" - «' . (6) 2g If the starting-point is below the origin, we should change the sign of Si. , , -. , .Li- • If the body is projected downwards, we should change the signs of v,vi,s and Si. We see that this is equivalent to simply chang- /I 94 KINEMATICS OF A POINT — KECTILINEAR TBANSLATION. [CHAP. T, ing the sign of g in all equations, leaving the signs of the other quantities unchanged. When the final velocity v is zero, we have from (2), for the time of rising to the highest point or the " turning-point,'''' T = — 9 ' For the time of rising to the highest point and returning to the starting-point we make s — Si = in (4) and obtain 2T = ^. 9 Hence, the times of rising and returning are equal. For body falling we have T = \ the minus sign denoting time before the start necessary to acquire the velocity Vi. The distance from the starting-point to the turning-point is found from ( 6 j, by making t? = 0, to be ^. 2g The distance ^ or - is called the height due to the velocity Vi or V ; that is, the distance a body must fall from rest in order to ac- quire the velocity Vi or t?. When the distances in rising and falling are equal we have s — Si = 0, or x^ = jr— , or tJi = V; that is, the velocity of return is 2g 2g equal to the velocity of projection. If the time of rising is less than T = — , the displacement s — Si is equal to the distance described. But if the time of rising is greater than 2' = — , the body reaches the turning-point and then falls from rest, and the entire distance described is distance described = ~ + -qCt — T)^ = — s = -^—- • (7) 2g^2^^ ' g 2g [Application of Calculas to the preceding Case.] — We can deduce the pre- ceding equations from our general equations (8) to (10), page 51. Thus from equation (9) we have, for acceleration directed downwards and therefore (— ), (page 50,) ^ ~ dt~ df ~ ^' Integrating, we have « = — = — gt-\- Const. CLt When < = 0, let « = -f «i , the (-|-) sign denoting motion upwards. Then we have Const. = Vi, and * = S = ^'-^*' • • • ^^> which is equation (1), page 93. Integrating again, s = Dii — ~gti ■\- Const. CHAP. I.] CALLING BODY— EXAMPLES. 96- Let 8 = + Si when t = 0, tlie (-|-) sign denoting distance upwards. Then we have Const. = s, , and 8-Si = Vit--gt\ (3> which is equation (3), page 93. From equations (1) and (3) we can deduce all the others page 93. The student should note especially that these equations have been deduced for body projected upwards or Vi positive. If we suppose motion towards the centre or downwards, we should have ©, negative. Also if the starting-point is below the origin, we should change the sign of Su In all cases we take g minus, as long as the acceleration is directed down- wards. EXAMPLES. Unless otherwise specified g = 32.2 ft.-per-sec, per sec. or 981 cm.-per-sec. par sec. All bodies supposed to move in vacuum. ., V v(i) A point moves with a uniform velocity of 2 ft. per sec. Find ij ' the distance from the starting-point at the end of one hour. " ^Ans. 7200 ft. Motion in a straight line. -• -(2) Two trains have equal and opposite uniform velocities and _ each consists of 12 cars of 50 ft. They are observed to take 18 sec. to pass. Find their velocities. Ans. 22.73 miles per hour. ^ Two points move with uniform velocities of 8 and 15 ft. per sec', in directions inclined 90\ At a given instant their distance is 10 ft. and their relative velocity is inclined 30° to the line joining them. Find (a) their distance when nearest ; (b) the time after the given instant cut ichich their distance is least. Ans. (a) 5 ft. ; (6) ^ j/S sec. (4) A body is projected vertically upwards with a velocity of 300 ft. per sec. Find (a) its velocity after 2 sec.; (b) its velocity after 15 sec; (c) the time required for it to reach its greatest height ; id) the greatest height reached ; (e) its displacement at the end of 15 sec; if) the space traversed by it in the first 15 sec ; {g) its dis- placement ivhen its velocity is 200 ft. per sec. upwards ; (h) the time required for it to attain a displacement of S20ft.* Ans. (a) 235.6 ft. per sec. upwards; (6) 183 ft. per sec. downwards; (c) 9.3 sec; (d) 1397.5 ft.; (e) 877.5 ft. upwards; (/) 1917.5 ft.; (g) 776.3 ft. upwards; (h) 1.13 sec. in ascending, 17.5 sec. in descending. {,5) A ball is projected upivards from a window half ivay up a toicer 117.72 meters high, with a velocity of 39.24 m. per sec. Find the time and speed (a) with which it passes the top of the tower ascending ; (b) the same point descending ; (c) reaches the foot of the tower. Ans. (a) 2 sec; 19.62 m. per sec; (ft) 6 sec; 19.62 m. per sec; (c) (4 -f 24/7) sec; 19.62 i^l m. per sec. * If the student will refer to the Examples, page 114, he will gain an idea, of the effect of the air in modifying the motion of falling bodies, and will better appreciate the delusive nature of all problems which ignore it. 96 KINEMATICS OF A POINT — RECTILINEAR TRANSLATION. [CHAP. I. ^>/L,(6) A stone is dropped into a well and tJie splash is heard in 3.13 kec. If sound travels in air with a uniform velocity of 332 meters per sec, find the depth of the well. Ans. 44.1 meters. ^7) If in the preceding example the time until the splash is heard ^ and the velocity of sound in air is V, find the depth. -:d Ans. Depth = -\ (Tg-^ V)- \/V{2Tg-{- V) / pjS) Show that a body projected vertically upwards requires twice as'long a time to return to its initial position as to reach the highest point of its path, and has on returning to its initial position a speed equal to its initial speed. (9) A stone projected vertically upwards returns to its initial position in 6 sec. Find (a) its height at the end of the first second, and {b) what additional speed would have kept it 1 sec. longer in the air. Ans. {a) 80.5 ft.; (J) 16.1 ft. per sec. (10) A body let fall near the surface of a small planet is found to traverse 204 ft. between the fifth and sixth seconds. Find the ac- celeration. Ans. 20.4 ft.-per-sec. per sec. (11) A particle describes in the nth second of its fall from rest a space equal to p times the space described in the (n — l)th second. Find the lohole space described. ga - Spy Ans. 8(1 - py (12) A body uniformly accelerated, and starting ivithout initial velocity, passes over bfeet in the first p seconds. Find the time of passing over the next b ft. Ans/jo( '^^2'- l) sec. [3) A ball is dropped from the top of an elevator 4.905 meters high. Acceleration of gravity is 9.81 meter s-per -sec. per sec. Find the times in which it will reach the floor (a) when the elevator is at rest; (b) ivhen it is moving with a uniform downward acceleration of 9.81 m.-per-sec. per sec; (c) when moving with a uniform down- xoard acceleration of 4.905 m.-per-sec. per sec. ; (d) ivhen moving with a uniform upward acceleration o/ 4.905 m.-per-sec. per sec. Ans. (a) 1 sec. ; (i) oo ; (c) |/3^sec. ; {d) y , I sec. (14) If Si , Si are the heights to which a body can be projected ivith a given initial vertical velocity at two places on the eartlvs sur- face at which the accelerations of falling bodies are gi and g^ respec- tively, show that Sigi = Stg-i. 5) A stone A is let fall from the top of a tower 483 ft. high. Ai the same instant another stone B is let fall from a window 161 ft. below the top. How long before A will B reach the ground f Ans. ( 1/6 — 2) V5"sec. ■CHAP. I.] FALLING BODY — EXAMPLES. 97 ^ u<re) A hall falling from the top of a toicer had descended e^feet when another was let fall at a point b below the top. Show that if they reach the ground together, tJie height of the tower is - "^ ' ft. 4a (17) If two bodies are projected vertically upwards with the same initial velocity F, at an interval of t sec, prove that they will meet at a height ^{-^--^. (18) Tivo stones are falling in the same vertical line. Show that if one can overtake the other, it ivill do so after the same lapse of time, even if gravity ceases to act. (19) Bodies are projected vertically doivnwards from heights hi , hi , hi with velocities Vi, v^, Va , and all reach the ground at the same moment. Shoiv that hi — hi hi — hi ha — h^ Vi —Vi Vi — Va Vs — Ui ' (20) Two points move in straight lines with uniform accelerations. Show that if at any instant their velocities are proportional to their respective accelerations, the path of either relative to the other will be rectilinear. (21) Upon the top of a toicer 200 feet high is placed a flag-staff of 26 feet ; a bullet is Let fall from the top of this flag-staff, and at the instant of its passing the bottom of it a stone is l^ fall from a win- dow 44 feet from the top of the tower. At what distance from the bottom of the tower will the bullet overtake the stone f Show also that this distance is independent of the value of g. In what tim^ after the dropping of the stone do they meet f In what time after the dropping of the bullet ? How far does the stone fall before meet- ing f Take acceleration due to gravity 32.16 ft. per-sec. per sec. r 44' Ans. Distance = 200 — 44 + ^— 4X26 = 137.385 ft. Time from falling of stone 1.07 sec. Time from falling of ballet 2.34 sec. The stone falls 18.615 ft. The bullet falls 88.615 ft. (22) A body falls a distance a from rest when another body is let fall from a distance a + b below the starting-point of the first. How far tvill the latter body fall before it is overtaken by the for- mer ? What is the time of fall of the latter body f --; time = 4/ ^ — . 4a r 2ag (§3) A body is projected upward with a velocity which would take it to the height a, and at the same instant a body is let fall from a distance above the point of projection of b. At what dis- tance below the latter point will the bodies meet f In what time f Ans. a; = -— ; time = 4/ ^ — . 4a r 2ag (24) A body is thrown vertically upward' with a velocity Vi. Find the time at which it is at a given height h in its ascent. Ans. , = ^_L±J^£ZZM. a The lower sign gives the time when the body is at the height h in ascend- ing, the upper in descending. 98 KIKEMATICS OF A POINT — KECTILINEAR TRANSLATION^. [CHAP. I» (25) A body is projected vertically upward and the interval be- tween the times of its passing a point whose height is h in its ascent and descent is 2t. Find the velocity Vi of projection and the whole time T of motion. 2\/(ff^ + 2gh Ans. Vi = ^gH'' + 2gh, T = 26) A body falling to the ground is observed to pass through eight ninths' of its original height in the last second. Find the height. 9 Ans. -^ = 36 ft. nearly. '-I^TM i7) A body falling under the action of gravity is observed to describe 144.9 feet and 177.1 feet in two successive seconds. Find g arid the time from the beginning of the motion to the first of the two seconds. Ans. g = 32.2 ft.-per-sec. per sec, t =4 sec. (28) A falling body is observed at one portion of its path to pass through nfeet in t sec. Find the distance described in the next t seconds. Ans. n -|- gf feet. (29) A body is projected vertically upwards with a velocity '6g. At what times wilt its height be 4g, and what will be its velocity at these times f Ans. It will be at the height 4g at the end of 2 sec. and again at the end of 4 sec. Its velocity at both these instants is g ft. per sec. upward at the end of 2 sec. and downward at the end of 4 sec. (30) Find the velocity with which a body must be projected up a smooth inclined plane., the height of which is h and length Z, to reach the top. Ans. The vertical acceleration is g. The component acceleration parallel to the plane is a = ^. Therefore « = ^2al — i^2gh, or the same as the ve- locity required to project the body to the height h. ^"'fsi) Find the time of falling down the whole length of a smooth inclined plane of length I and height h. Ans li/ —• ^^^ constant height h the time is directly as the length. r gh ^"^2) Find the speed attained by a body in falling down a smooth inclined plane the height of which is h and length I. Ans. \^2gh, the same as in falling through the height 7t. ^•'^tSS) A body is projected down a smooth plane the inclination of which with the horizontal is 45°, with a velocity of 10 ft. per sec. Find the space described in 2i seconds (g = 32). Ans. 95.7 ft ^^34) A locomotive starts down a smooth incline with a velocity of 7i miles an hour. If the ratio of the height to length is gx^, find the space traversed in two minutes {g = 32). Ans. 2472 ft. CHAP. I.] ACCELERATION INVERSELY AS SQUARE OF DISTANCE. 99 r Acceleration Inversely as the Square of the Distance from a Fixed f Point— Motion Rectilinear— Force Attractive.— This is the case of i a body at a great distance from the earth, under the action of the force of gravity, since in such case the acceleration is towards the centre of the earth and varies inversely as the square of the dis- tance from the centre of the earth.* Let si be the initial distance from the centre of acceleration, the velocity at this point being v^ , and s the distance to any other posi- tion at which the velocity is v. If the acceleration is towards the centre or force attractive, it is negative (page 50). Let a' be the known acceleration at a distance r', and a the accel- eration at any distance s. Then we have a:a'::r"': s", or a = ar Thus for instance, in the case of the earth, a' is gr, and r' is the radius of the earth at the locality where g is known. -«• We have then, if r is an indefinitely small time, for acceleration towards the centre = a = — (1) The mean velocity for an indefinitely short time is V + Vi , and the distance described in this time is S — Si = Multiplying by (1), -r, or V + Vi = 2(s — 8i) 2a'r'\ V^ —Vi'= -— (S — Si). s" But if the time is indefinitely small, s" will equal ssi , and hence v^ _ Vi" = - 2a'r''(- - -V \8i sj .... (2) or v^ = Vi^ - 2a'r''l- --\. If the body falls, Si is greater than s and the last term becomes essentially positive. If the body is projected upwards, Si is less than s and the last term is essentially negative. Equation (2) holds good, then, without change in either case if the acceleration is towards the centre, or force attractive. It also holds for any path * This is the ' law of universal gravitation " as discovered by Newton. It is also known as the law of the inverse squares. It is regarded as rigidly true for every particle of matter acting upon every other particle. But, as we shall see, it is not rigidly true for bodies of finite dimensions acting upon similar bodies, unless those bodies are homogeneous spheres or spherical shells. The earth is not a sphere and is not homogeneous. Therefore it is not rigidly true that it attracts external bodies with a force inversely as the square of the distance from the centre. The deviation from this law for bodies at great distances is, however, insensible. 100 KINEMATICS OF A POINT — RECTILINEAB TRANSLATION. [CHAP. I. straight or curved if s, a, , r' are measured along the path, and a' is the tangential acceleration or rate of change of speed at distance r\ and V and Vi are the speeds final and initial. If the acceleration is away from the centre, or force repulsive, we should have the sign before the last term (+) instead of (— ), or the sign of a' is changed in (1) and (2). If the body falls from rest from a distance Si , we have from (2), by making Vi = 0, for a body falling from rest. v' = 2a'r"'(- - -] falling. (3) If the body is projected upwards and the velocity v is zero at the height s, we have from (2), for a body projected upwards to a distance s, Vi'^ = 2aV"( 1 rising (4) Cor. 1. If the distance Si is infinite, we have from (3) the velocity acquired in falling from an infinite distance, v = - y- 2a'r'^ and from (4) the velocity of projection in order to go to an infinite distance is , ./2a'r" vn In the case of a body attracted by the earth we have a' = g. If then in the first case s = r' and in the second case Si = r', we have for the velocity acquired in falling to the surface of the earth from an infinite distance, or the velocity necessary to project a body to an infinite distance from the surface of the earth, in vacuo, V — Vi — T V2gr'. If we take gr = 32i ft.-per-sec. per sec. and the mean radius of the earth 3960 miles, we have ^Q^ X 3960\i ^ „ ^^ ., v = Vi = [ — ^--^- — I = T 6.95 miles per sec. \ 5280 j The ( — ) sign for falling and the (+) sign for upward projection. Cor. 2. If we put equation (3) in the form V' = 2gr"l-— ^ \ ss we see at once that if Si — s is a small distance compared to s, so that the entire fall takes place near the earth's surface, ssi will be practically equal to r"* and we shall have u" = 2g{Si — s). This is the same formula as for uniform acceleration g towards the earth, the initial velocity being zero (page 93). [Application of Caleolas to the Preceding Case.] — We can deduce the preced- ing results from our general equations (8) to (10), page 51. Thus for acceleration towards the centre we have, as before, _ dv _d^s _ a'r'^ ^. dt dt^ Civil & Mechanical Engineer. SAN FRAiN CISCO, CAL, CHAP. I.] ACCELERATION IKVEESELY AS SQUARE OF DISTANCE. 101 For acceleration away from the centre we should have (-f-) instead of (— ) (page 50). ds Multiply by ds, both sides, and then, since t- = ■», we have vdv = — a'r'^ds Integrating, we have <' -J«^ a'r" -f- Const. When s = Si , let v = -\-Vi for body projected upwards and v = —Vi for body projected downwards. Then in both cases Const. = —-7 , and in 2 Si both cases «» = ®,» -2a'r'^(~ --V (2) These are the equations (1) and (2) of the preceding Article. They hold, as we see, for motion towards or away from the centre, provided the acceleration is towards the centre. For acceleration away from the centre we change the sign of a'. These equations also hold for any path, straight or curved, if s, Si , r' are measured along the path, and a' is the tangential acceleration or rate of change of speed at distance /, and d and Vi the speeds final and initial. If the initial velocity is zero, we have for a hodjfaUing from rest ds «»^ 2ay''{---\. (3) Since ® =; — , we have from (3) -f = -A-(^i). where we take the (— ) sign for the radical to denote motion towa/rds the centre S (page 44). This can be put in the form sds ,^ , /%a'r'^ — = at 4/ . 4/siS — s" ^ *• To put this in a form convenient for integration, add and subtract ^«i to the numerator of the first term. We then have I \Sx—S — 2« Integrating, we have 2 i/s.s - 8» 2 /« SxS — «* r Si «.* Si . _i 2s (siS — «*)^ — TT versm — ^ ' 2 « s _ /2a'r'y. t + Const. (4) Let t iCSi when s = Si, then Const. = ~. Hence* for the time of falling -,2s _i/, 2«\ _,/2s ,\ * We have n — versm — = ;r — cos 1 1 1 = cos ( 11. From trigonometry, 2 cos« y — 1 = cos 2y. Let 2y = cos" ' f -^ 1 j . Then cos 2v = - — 1 and cos" w = — , or jr = cos^'i/— and 2y = 2 cos"'4/ - . \ Hence 2 cos . _i2s versin — Sl 102 KINEMATICS OF A POINT — RECTILINEAR TRANSLATION. [CHAP. I. [ from rest we have for acceleration towards the centre ; If we had taken motion away from the centre, we should have obtained, ' instead of (4), ^ — (siS — «") + ^ versin — = I ) < -|- Const, .rr C 8i \^ 8i J Let t = 0, when 8 = 0, and Const. = 0, and we have I Equation (6) applies to a body projected upwards to any height s, Si being the height at which it would come to rest. If we make s = Si in (6), or « = in (5), we find the time of reaching the turning-point in rising or reaching the centre in Tailing from rest AT 2 ^^'r'^ $ / CHAPTEE II. SIMPLE HARMONIC MOTION. MOTION IN RESISTING MEDIUM. ^ Simple Harmonic Motion. — The motion of a point moving in any path in such a manner that the tangential acceleration is directly proportional to the distance, along the path, from a fixed point in the path is called simple harmonic motion. Such motion may be rectilinear or curvilinear. The vibrations of such bodies as a tuning-fork or a piano-wire are approximate examples of such motion, and hence the term " harmonic." The vibrations of an elastic body, such as the air, are examples of such motion. It is also, as has been stated (note, page 92), the motion of a body under the action of gravitation, within a homogeneous sphere, as it can be shown that in this case the acceleration due to gravity is proportional to the distance from the centre. The motion of the piston of a steam-engine when moved by a crank and connecting-rod approximates the same motion if the rotation of the crank is uniform, the approximation being closer the longer the connecting-rod. This wUl be evident from the fol- lowing Article. Simple Harmonic Motion in a Straight Line — Force Attractive. — Let a point M move with uniform speed in a circle of radius CM = r. Then the acceleration fn is always direct- ed towards the centre and equal to fn = rm", where go is the constant angular velocity (page 76). The projection of /71 upon the diameter A ^ CA is fijV cos MCP. But r cos MCP is the distance CP=s of the projection P of M upon the diameter CA. Therefore the projection of fn upon the diameter is a = m's, or, since go is constant, a is directly proportional to the distance \ CP = s. The motion of P is therefore harmonic. \ If then a point M moves with uniform speed in a circle, its pro- ^Jection P itpon any diameter moves with harmonic motion in the ^__ii/am€fer, the centre of acceleration being the centre of the circle. , Let a' be the known acceleration in the Hne AC of P at a given I /„T ^^^ i -distance r' from the centre. Then a' = mV and 00 =a/ ^ , and the i ' * 103 104 i \ KINEMATICS OF A POIIfT — TRAN8LATI0K. [CHAP. II. speed of M is roo = r|/?-. The projection of this speed on the diameter is 4/?^ r cos CMP= MP a/—. ' V r -V.' But MP = |/r" — s' ; hence we have for the velocity of the point P in the lino ^C at the distance s = CP from C, = ^,{r^-8^). (1) or, if Vi is the initial velocity, s\ where a' is the know n acceleration of P at a given distance r' from Thus the point P starts from rest at the distance s = r from C. The velocity increases as the distance s decreases, till P arrives at the centre C where the velocity is a maximum and equal to V = r |/ "-. Then the velocity decreases and finally becomes zero when P arrives at A' at the distance s = — r on the other side of C. Equation (1) holds good for motion towards or away from the centre if the acceleration is towards the centre, or force attract- ive. It also holds for any path straight or curved if s, r, r' are measured along the path and a' is the tangential acceleration or rate of change of speed at distance r', and v and v' the speeds final and initial. Cor. 1. Since the uniform speed of ilf in the circle is ri/— , the* r ^> time occupied by P in passing from A to A' and back to A is But if a is the acceleration at any distance s, and /„ is the ac- celeration at the extreme distance r, we have for harmonic motion r' _ r _s _ JL_ a' ~ fn a <»'" Hence the time T of a complete oscillation is GO 'a \ a.] \ The time T of a complete oscillation depends therefore only upon the constant ratio — = , and is independent of the range r or amplitude of the oscillation. For this reason the oscillations are said to be isocbronous, or made in equal times, no matter what the range or amplitude. Cor. 2. Since the motion of a body under the action of gravity in a homogeneous sphere is harmonic (page 92), if we put g for a' and let r' be the mean radius of the earth, we have from (1) the motion of a body falling under the action of gravity towards th& CHAP. II. J SIMPLE HARMOKIC MOTIOlf. 105- P centre of the earth in a well or shaft, assuming the earth to be a homogeneous sphere and neglecting resistance of the air. In such case (1) becomes u" = %(r + s){r - 8). r If the fall takes place for a short distance compared to r' and near the surface, we have r + s practically equal to 2?'' and hence v^ = 2g(r — s), which is the same as for uniform acceleration g, the initial speed being zero. We obtained the same result (page 93) for a body external to the earth. The equations of page 93 hold good, therefore, in all practical cases, whether the fall takes place above the earth or within the earth, neglecting resistance of the air. Amplitude — Epoch — Period — Phase. — The range r = CA = CA' on either side of the centre of acceleration, in harmonic motion, is called the amplitude. A complete oscillation is from A to A' and back to A. The time of an oscillation, as we have seen, is independent of the amplitude. From A to A' or A' to A is a vibration. A vibration is half an oscillation. The time of a vibration is half that of a complete oscillation. If Pi is the initial position from which the time is counted, or the position of P at zero of time, the time of passing from A to Pi is called the epoch. The epoch may also be defined with reference to the auxiliary circle, as the angle ACMi in radians. This is the epoch in angular measure. The epoch in angular measure is then the angle described on the auxiliary circle in the interval of time defined as the epoch. The epoch locates the position of P at zero of time. The entire time which elapses from any instant until the moving point again moves in the same direction through the same position is called the period. The time from Pi to A\ then back through Pi to A, and finally back from A to P , is a period. It is evidently the time of a complete oscillation from A back to A. That fraction of the period which has elapsed since the moving point P last occupied A is called the phase. Measured on the circle, it is the ratio of the angle ACM radians to 2;r radians. Uie phase locates the position of P at any instant. It therefore varies with the time or with the position of P. The phase at zero of time, then, multiplied by 2?^ radians gives the epoch in angular measure, and multiplied by the time of an oscilla- tion gives the epoch in time. [Application of Calculus to Harmonic Motion. — We may deduce tbe results obtained for simple harmonic motion (page 104), as well as others, from the general equations (8) to (10), page 51. F We have, as before, a = —.8 (page 104). r For acceleration away from the centre we have a positive, for acceleration towards the centre a negative (page 50). 106 KINEMATICS OF A POI^TT — TRANSLATION. [CHAP. II. I. Acceleration towards the Centre. — In this case we have _ dv _ cPs _ a' ds Multiply both sides by da and then, since — = v, we have •vdv = -sds. r Integrating, we obtain ^' = -^ + Const. . .^ . . , . . (a) ^ lave then ^ c • ^ ^"^^ / V ^ J^ When « = r let « = ®i. We have then ^ z ' „— - '*' •♦ r f ^ aj • s V ' — "^ Const. = ui» + ^, ^' and hence «* = ^'^ + 7(^' - «*) (1) If the initial velocity «i is zero, this becomes (2) which is the same as equation (1), page 104, already obtained, for initial velocity aero and range r. ds Since « = — , we have from (1) -» = :j7 = ± i/ «i + -(^ - «"). dt r r where we take the (+) sign for motion away from the centre and the (-) sign for motion towards the centre (page 44). This can be written _ ds , /a' = = ±y-,dt (8) r a If we integrate this between the limits of t and < = when s = r, we have s ^ . /a' , . _i r Y ' a' fa s = r cos t y -; ± Vi i/ — sint y -^ (4) * Let ^ = ±t i/ -„ B= sin- r a «i' Then — - — = sin {A -\- ^ = sin A cos B + cos A sin B. r a But sin 5= — - , and cos 5= 4/I — sin' B. Substituting these /— — T— A values and reducing, we obtain equation (4). ^^ ^at — • ]/ ^ i?Cv CHAP. II.] SIMPLE HARMOKIC MOTION. 107 If motion is towards the centre, we take the ( — ) sign ; if away from the centre, the (+) sign. If ©1 is zero, or there is no initial velocity, we have « = r cos t ^^ ' = \/l' If we make u = in (1), we have for the amplitude = R= ± |/r» a This reduces to ± r when ®i = 0. If we integrate (3) between the limits of t and < = when 8 = B, that is, if we count the time from the end of the amplitude where Vi = 0, instead of from S = r, we obtain and hence s = Bcosty -;; = 4/^ -1 s ^^^ B' (7) If in (5) .we make « = — r or in (7) make s = — B, we have in both cases for the time of a vibration, t 4/ -7-1 and hence for the time of a complete oscillation T — 2it 4/?. Therefore tTie time of oscillation or vibration is not affected by the initial velocity. All these equations (1) to (7) hold for motion either towards or away from the centre, provided the acceleration is towards the centre. They also hold for any path, straight or curved, provided r*, r and s are measured along the path and a is the rate of change of speed at the distance r'. II. Acceleration Away from the Centre. — In this case we have the accelera- tion positive and hence dv , a' dt ' r ds Multiplying by ds, we have, since — = «, Integrating this, we have -7- 4- Const. When « = 0, let c = Oi. Then Const. = Vi^ and r (8) From (8) we have ds . / dt ('J «,'+^A ^t^ '" where we take the (+) sign for motion away and the (— ) sign for motion towards the centre. We can put this in the form 108 KINEMATICS OF A POINT — TRANSLATION. [CHAP. II. way ai form _v^. ^^ If we integrate this between the limits of t and t = when « = 0, we have logn («+ j/«'' + ^,«.») = ± <|/^ + logn vY^,. Hence (9) 2 y a'[ /a' • where e is the base of the Naperian system of logarithms. EXAMPLES. g = 32.16 ft.-per-sec. per sec. or 980.23 cm.-per-sec. per sec. (1) If the radius of the earth is 6370900 meters and the accelera- tion of gravity 9.81 meters-per-sec. per sec., ivhat should be the valtie of a'r'^ in eq. (2), page 99, «/ s and Si are given in kilometers f Ans. 398171.88 cubic kilometers-per-sec. per sec. (2) A body falls to the earth from a point 1000 miles above the surface. Find its speed on reaching the surface, neglecting resist- ance of the air and taking the eartKs radius 4000 miles. Ans. ® = 3.12 miles per sec. ^""^^ (3) In the last example find the distance from the earth's surface ' ■" when the speed is 2 miles per sec. Ans. 535.2 miles. (4) With uihat speed must a body be projected vertically at the earth's surface so that it may never return f (Assume the earth to ^^^ have no atmosphere and not to be rotating.) ....-"^'■v Ans. The speed is the same as that which a falling body would have fall- I'^J^ ing from an infinite distance, or « = 6.95 miles per sec. ,,r^ r^ i (5) At what point on a line joining the centres of the earth and ' ▼ ^'moon would the rate of change of speed of a body be zero f (At the *>• >- yi^ moon's surface a = 5.5 ft.-per-sec. per sec; radius of moon 108O i", ■^}^ miles : distance between centres of earth and moon 240000 miles.) V"^^. Ans. Let a; = distance of point from earth's centre and Xi from moon's ^9' centre, and i2 earth's radius, r moon's radius. Then x-{-Xi =240000, and , \.H ^ "il ^ ^J^i . Hence x = 215893 miles. ^ ar," :r i (G) A point whose motion is simple harmonic has velocities 20 and 25 ft. per sec. at distances 10 and 8 ft. from the centre of ac- celeration. Find (a) its period, (b) its acceleration at unit distance from centre .^ ,,.^.., /->o.'^ LaT^^^) r i /-,/> -CHAP. II.] TRANSLATION — EXAMPLES. 109 Ans. We have 625 = -,{f — 64) and 400 = ^(rj" - 100). Therefore T T /a 15 27r Ait a's 325 y -. = -^^^V^^^==~-^ = -^s^- a = - -r- = - -5« = - 6.25 ft.- ^ r , b /^ r 36 J* per-sec. per sec. fi (7) 7%e period of a simple harmonic motion is 20 sec. and the ^ maximum velocity is 10 ft. per sec. Find the velocity at a distance I . of — ft. from the mean position. Ans. — —z = 20 sec, therefore -7 = -r—-. Where « = 0. ri = dt. /^ ' ff *^ r Hence e" = ^(^-;^- ^1 or « = 8 ft. per sec. / ^ -^, (>^ ^ / '<^S) A point moves from rest towards a fixed point 10 meters dis- ^ '■ taut, its acceleration being everywhere 4 times its distance from a "^ ^' fixed point. At what distance will it have a velocity of 12 meters per sec. f —-^ Ans. 8 meters. (9) Find the mean speed of a point executing a simple harmonic motion during the time occupi&l in mx)ving from, one to the other extremity of its range, its maximum speed being 5 ft. per sec. Ans. The distance is 2r. The time — -=. The mean speed ^ ^^ Y r' When 8 = 0, we have 25 = —,r^, or r /l/ -, = 5. Therefore mean speed is - — r r r It ft. per sec. (10) If The the period and a the amplitude of a simple harmonic mixtion, and if v be the velocity and s the distance from the centre at a given instant, show that (11) A point oscillates about a centre, its acceleration being pro- portional to its distance. Show that the ratio of its maximum ve- locity to the square root of the excess of the square of its maximum velocity over the square of the velocity which it has when at a given displacement from the centre is equal to the ratio of its maximum displacement to the given displacement. (12) A point has a simple harmonic motion whose period is 4 trem. 12 sec. Find the time during which its phase changes from — to- of a period. Ans. 21 sec. [Body Projected in a Besisting Medium — Acceleration Proportional to the Square of the Velocity— Motion Bectilinear.]— When a body moves in a resisting ( //6^ 110 KINEMATICS OF A POINT — TRANSLATION. [CHAP. II. medium such as air or water it loses velocity or Las a minus acceleration, which is usually assumed to vary as the square of the velocity. We have then « = dr = -'^' ^i> where c is a constant depending upon the shape and dimensions of the body and the density, or mass of a unit of volume, of the body and medium. This constant is called the coefficient of resistance. For instance, for a sphere, if d is the diameter and A the density of the medium and 8 the density of the body, we have, as is proved in Vol. Ill, Kinetics, page 61, _£ J^ ■^e can put (1) in the form ^* ^^ • cdt = - -r. Let B = z>i when t = and integrate, and we have ^B 151/ C \V Vi) ' ''-T^t (^> From (2) and (3) we can find the time for any velocity or the reverse. Since « = -T-, we have from (3) Integrating and making « = when ^ = 0, we have «= — logn(l + CTjQ; ■ . (4) c or using common logarithms, 2.302585. .. , ^^ _ « = Iog(14-CB,0 (5) From (4) we have t = ^- -, (6) where e — 2.718282 = base of the Naperian system of logarithms. From (5) and (6) we can find the distance for any time or the reverse. From (6) we have coxt = e"* — 1, and substituting this in (8) we have •0=^ (7) or 1 -w «=-logn-; (8)' or using common logarithms 2.302585 , t), * = — 7— ^"^7 (^> From (7) and (9) we can find the velocity for any distance or the reverse. From (8) we see that when the velocity becomes zero the space traversed is infinite, and from (2) the time is infinite. Although then the velocity dimin- ishes as the time increases, as we see from (3), it cannot become zero in any finite distance or time. r CHAP. II.] RESISTIKG MEDIUM. HI [Body Falling under the Action of Gravity in a Kesisting Medium.]— Let /be the uniform acceleration due to any attractive force. In the case of ffravitv we have* ° ■' (T^^ w^TaLl gJ^ ^ l ! ^ l-hri (l epei fcy-o p ma ss-of-a uuit vulume ' O f t be - m ii dlum and i S -isHhe (teft>;ity of the body. "The acceleration / acts away from the starting-point and tlu' ntaidution cv' acts towards the starting-point. If then we take this point as origin, we have W^-^-'"' (1) where c is the coeflBcient of resistance and has the same value as in the preced- ing Article. Let k be that velocity for which the reta,rdation is equal to/, so that/= cA'^ Then c = p, and equation (1) becomes dv f We can write this in the form Integrating, we have :,. ^* <^« When < = 0, let t> = •»!. Then Const. = -~. logn ^l^i. Hence _ k {k-^v){k- «.) '-W^''^ ik-v){k + vS ' <*> or using common logarithms, _.. — -- "> _ 2.3Q2585A; (*+»)(* - ®i) ' -^sT"^"^ (A -. )(*+.,) !^ From (4) we have, if e = 2.718282 = base of Naperian system of logarithms. V = P'—e^ + i From (5) and (6) we can find the time for any given velocity or the reverse. * As we shall see hereafter, the mass of a body multiplied by g gives the weight of the body, that is, the force of gravity. It is also a well-known fact that a body immersed in a fluid has its weight diminished by the weight of an equal volume of the medium. If then V is the volume of the body, VSg is its weight in vacuo and V^g is its loss of weight due to the medium. Hence VSg — V.^g is the weight when fmmersed. Since VS is the mass, ^ ^ rdf= VSg - r^g, or f = gll- ~\ . U^ ^ /<:> f 112 KINEMATICS OF A POINT — TEANSLATION. [CHAP. II. Since «= -sr* we have ^ / -u 1 v and therefore from (2) we have "^ A ^' ^'^ iising common logarithms, ^./ U\ /^) ^ «^ - ^ 2.302585A;' , k^ - ^,^ ,^^ K. k,^ [i, ^-./, ^=— 2r-'°^-*^-:^ ^'^ '<^<;/^ '^ From'' (7) we have _ ^ «« = A;» - (^2 - B,s)e *' , (9) ■where 6 = 2.718282 = base of the Naperian system of logarithms. < From (8) and (9) we can find the distance for any velocity or the reverse. y^S/V" I We see from (9) that when s is great, v approaches k, and k is the limiting value of V. If the initial velocity is zero or less than k, v will continually ap- proach k, but can never exceed k. If the initial velocity is greater than k, v will diminish continually down to k and can never become less than k. . [Body Projected Upwards under the Action of ~ Gravity in a Kesisting Me- dium.] In this case we have, taking k as before, since /is negative and the resist- ance is negative, '' = t--f-i^ « which can be written dt = 35- 1-5— — 5. J «■'-(-«■' Integrating and determining the constant Ijy the condition that when t = 0, « = ®j = initial velocity, we have We have also as before -^ = -t— , and therefore, from (1), _ k^ vdv ^^ ^^Integrating this, and making « = 0, when ® = «i , we have 7I - /ijL^li/fMf Ipiritl: 2.302585A;2 . k^-\-'c,^ ■ * = — 2r" '^ "^M^ ^^^ The time in which the velocity becomes zero and the body reaches the turn- ing-point is, from (2), T=j'---% (5) d the corresponding value of s is, from (4), - 2.302585A;' . T , ®.n CHAP. II.] RESISTING MEDIUM, 113 At the end of the time T the body begins to return and falls from a state of rest, or Vi = 0. We have then from the preceding Article, making Vi = 0, , _, 2.302585^ , k+v _ and 2.302585A;2, k^ A-.= -^^log^^--^, . ...... (8) Let u be the velocity with which the body returns to the starting-point. Then putting « = in (8) and v = u, we have or substituting for h its value, t = ~r <») Hence ^-^=F <10) We see then that u is less than Vi, or the body returns to the point of pro- jection with a velocity less than the velocity of projection. Values of —r and c. — For motion in a resisting medium, under the action of gravity, we have/ = fl'fl ~ t)> where ^ is the density of the medium and 5 that of the body. ft For iron in water we may take ^ = 7.2. " " " air " " " ^=5983.28. " mist or rain in air " " " - = 813.82. " lead in water " " " j = 11-35. St ♦' " " air " " " 2 =9423.61. The coefficient of resistance for a sphere (Vol. Ill, Kinetics, page 61) is - L - J£. *^~Jc'~ 16<Jr' where r is the radius of the sphere. For a cone we have _ 3/Jr'' where r is the radius of the base and h the height. If the cone terminates in a cylinder of length I, we have _ Sdr" ^~2d{3l+h){r' + h')' 114 KINEMATICS OF A POINT — TRANSLATION. [CHAP. II. EXAMPLES.* g = 32.16 ft.-per-sec. per sec. (1) A lead bullet 1 inch in diameter is projected vertically ivith a velocity of 2000 ft. per sec. Find (a) the time of accent with and ivithout resistance of the air ; (b) the distance to which it ascends with and without resistance of the air ; (c) the velocity and time of return with and without resistance of the air. Ans. We have in this case % = 9423.61, ki = 67338.1852 and k = 259.49 ft. per sec, / = 32.1556 ft.-per-sec. per sec. (a) The time of ascent in vacuo is 62.19 sec. _ . _, 259.49 ^ , 2000 .. ._ '^ "■'' ^= 3Tl556*^" - '259:49 = 1^-^" '^- (b) The height of ascent in vacuo is 62189 ft. T . , 2.302585 X 67338.1852 / 4000000 \ „-„„ r. In air h = ^^ ....„ log 1 4 1 = 8583 ft. 33.1006 ^ \^^ 67338. 1852 y (c) The velocity of return in vacuo is 2000 ft. per sec. 4000000 In air fj.'^ = ■ 4006{)0(r' ^^ ^ = ^57 ft. per sec. 1 + 67338.1852 (<?) The time of return in vacuo is 62.19 sec. ^ . ^ ^ 2.302585X259.49, 259.49 + 257 „, ,„ In air « - 7= 2x32.1556 ^^^ 259.49 - 257 = ^^'^^ '^*'- (2) A lead bidlet 1 inch in diameter is let fall in the air. Find the velocity at the end of t = 1 sec., 2 sec, 3 sec, 10 sec, 20 sec, with and without the resistance of the air. Ans. We have k = 259.49 ft. per sec; /= 32.1556 ft.-per-sec per sec; (^ \ k\e * — 1/ e = 2.718282; and from eq. (6), page 111, making «i = 0, v= — ^ e '' + 1 Let t = 1,2, 3, 10 and 20. and we have d = 31.98, 63.03, 92.33, 219.3 and 255.86 ft. per sec; while in vacuo we would have® = 32.16, 64.32, 96.48, 321.6 and 643.2 ft. per sec. (3) In the previous example what is the greatest velocity the bullet can attain f Ans. k = 259.49 ft. per sec. As we have seen, this velocity is attained quite early, after which the velocity is uniform. (4) An iron cannon-ball 1 ft. in diameter is projected verticaUi/ upwards in the air with a velocity of 2000 ft. per sec. Find (a) the time of ascent; (b) the distance to which it ascends ; (c) the velocity with which it returns ; (d) the time of return. Ans. We have in this case -, = 5983.28, k^ = 513040, k = 716.268 ft. per 8ec.,f = 32.1546 ft.-per-sec. per sec. * An examination of these problems will give the student an idea of the- effect of the air in modifying the motion of a falling body and enable him to realize the inaccuracy of neglecting it. CHAP. II.] BESISTINQ MEDIUM — EXAMPLES. 115 (a) The time of ascent is T — 27.32 sec; in vacuo, 63.19 sec, (See Ex, (1).) (6) The distance of ascent is h = 34680 ft.; in vacuo, 62189 ft. (c) The velocity of return is u = 6T4.3 ft. per sec; in vacuo, 2000 ft. per sec (d) The time of return is < — T = 88.9 sec; in vacuo, 62.19 sec (5) An iron cannon-ball 1 ft. in diameter is let fall in the air. Find the velocity at the end of t = 1 sec, 2 sec., 3 sec, 10 sec, 40 sec, 60 sec. Ans. We have k = 716.268 ft. per sec.,/= 32.1546 ft.-per-sec per sec. and Vt = 0. Hence from eq. (6), page 111 : For f = 1 2 3 10 40 60 sec. V - 32.12 64 95.85 301.5 677.8 709.7 ft.-per-sec. In vacuo we would have ® = 32.16 64.32 96.48 321.6 1286.4 1929.6 ft. per sec (6) In the previous example, what is the greatest velocity the can- non-ball can attain f Ans. k = 716.268 ft. per sec. And this is attained in little more than a minute. (7) In Example (5) what are the distances passed through f Ans. From eq. (8) we have, making Ci = 9 and taking the values of « already found. For t=l 2 3 10 40 60 sec. 8 = 16.03 64.29 144.2 1561 18000 31992 ft. In vacuo we would have s = 16.08 64.32 144,73 1608 25728 57888 ft, (8) A lead shot i inch in diameter is let fall in the air. Find the greatest velocity it can attain, and the velocity and space traversed in t = 1,2, 3, 4, 5, 6, 7 and 8 sec. Ans. We have -|- = 9423.61,/= 32.1556, k^ = 8417. Greatest velocity = A; = 91.7 ft. per sec. For t = 1 V = 30.89 2 55.49 3 71.76 4 81.23 5 6 86.35 89.01 7 8 sec. 90.35 91.03 ft. per sec In vacuo, v = 32.16 64.32 96.48 128.64 160.8 192.96 225.12 257.28 ft. per sec » = 15.67 60.27 123.56 199 283.3 370.7 458.1 545.5 ft. In vacuo, 16.08 64.32 145.72 257.3 402 578.8 787.9 1029.13 ft. (9) What is the greatest velocity a rain-drop i inch in diameter can acquire, falling in the air f Ans. We have — p = 813 82, / = 32 12, A; = 27 ft. per sec. (10) An iron cannon-ball 1 ft. in diameter is let fall in water. Find (a) the greatest velocity it can attain; (6) the final velocity and the space passed through in t = 1, 2, 3 sec Ans. We have -^ = 72,f= 27.7. Therefore, (a) k = 23.06 ft. per sec (6) For < = 1 2 3 sec. ?)= 19.22 22.68 23.02 ft. per sec. s=1138 32.8 53.94 ft. We see that the maximum velocity is reached in about 3 sec. 116 KINEMATICS OF A POINT — TRANSLATION. [CHAP. H, In the preceding examples it is assumed that the density of the medium is unchanged, and that the acceleration of gravity is con- stant. Near the earth's surface both assumptions are practically true. We have also assumed that the acceleration varies as the square of the velocity. Experiments would seem to indicate that this is not strictly accurate. The effect of resisting media upon the motion of projectiles is therefore best taken account of by means of empirical formulas based upon experiment. We have given the preceding examples in order to call attention to the fact that the influence of the medium, even of the air, is such as to very materially modify the results of the formulas of page 93, which hold good only in vacuo and are not even approximately true except for large and heavy bodies for the first few seconds of fall. The examples of page 95 are therefore devoid of practical value except under such limitations. For very ^reat distances the density of the air and the accelera- tion of gravity are not constant, so that our present assumptions are then no longer in accord with fact. Vi^ /4V/3 ^ AV- /\Mr^^^^ pcj. CHAPTER III. TEANSLATION IN A CURVED PATH DIRECTION OP ACCELERATION CONSTANT. PARABOLIC MOTION. PROJECTILE IN A RESISTING MEDIUM. MOTION OF Curved Path, — When a point moves in a path such that the di- rection of the acceleration coincides with the direction of motion and does not change, the motion is rectilinear, no matter what the law of variation of the magnitude of the acceleration may be. Such ^ motion we have discussed in the preceding Chapters \ If the direction of the acceleration, however, does not coincide \ with that of the motion, then, whether it is constant in direction , and magnitude or not, we have motion in a curved path. When a rigid body composed of many points moves so that ' every straight line through any two of its points remains parallel to itself in all positions of the body, it has a motion of trans'lation only, and we may treat the body as if it were a point. For motion in a curved path the differential equations of page 81 apply. [ Uniform Acceleration Inclined to Direction of Motion.— If the ac- \ celeration is uniform, that is, constant in magnitude and direction, V its component in any given direction is uniform, and the equations for rectilinear motion, page 93, apply to the component motion in that direction. The most common case of curvilinear motion under uniform ac- ' celeration is that of a body projected with any given velocity in any given direction at the surface of the earth, neglecting the resistance of the air. In such case the acceleration due to gravity is practically uniform and equal to g ft.-per-sec. per sec. ; Let the initial velocity of projection Vi of the point P make the I angle APB — cxi with the hori- ^^ i zontal. \ Let the co-ordinates of any \ point M of the path or trajectory \ he PB = x and BM = y. \ Let the angle MPB = S. Let \/be the uniform vertical accel- Wation (in the case of gravity/ V9O: -^- — TTwe make Op parallel and equal to the velocity tJi at P, and ^ Om parallel and equal to the ve- kA locity V at any point M of the "^ trajectory, then ^»i = ft is the integral acceleration for the time t during which the point passes , . ,, ^ ^ from P to M, and the straight line pb is the hodograph (page 52) 117 118 KINEMATICS OF A POIKT— TRANSLATIOIS". [CHAP. HI. Is ;^ for motion from P to the point C, where the velocity is horizontal, and Ob is the velocity at this point. The velocity of p in the hodograph is the acceleration in the path (page 52). Therefore the point p moves with uniform velocity / from p to b, while P moves from P to C We see at once that the horizontal component of the velocity Vi is Ob = Vx, or Ob = Vx = Vi COB ai (1) The horizontal distance passed over in any time t, while the point P moves from P to if, is then PB = x, or PB = X — Vit cos ai (2) The vertical component of the velocity Vi is bp = vi sin ai up- wards. But the acceleration / is downwards Hence the vertical velocity at the end of the time t is bm = bp — pm = Vy, or bm — Vy= Vi sin ai — ft (3) The vertical velocity at the beginning of the time t is Vi sin ai . The mean vertical velocity during the time t is then 2Vi sin ai ~ ft . 1 ^ ~ = Vi sin ai — —ft. The vertical distance passed through in the time t is then BM = y, or BA-AM=BM~y = Vit8in.ai-~ft' (4) /* If we combine (2) and (4) by eliminating t, we have for the equa- tion of the tra^ctory i/ = a;tanari— / .^ (5) 'ivi^ cos^ ai This is the equation of a parabola. I The time of reaching the highest point C is the time of describ- / ing the vertical distance DC. Call this time Tv . Since at this point \ the vertical velocity is zero, we have, by making % = in (3), ™ _ ■Wi sinai Tv-—j—. If we substitute this for t in (2) and (4) we obtain the co-ordinates of the vertex C of the parabola, Vi'-' sin ai cos ori Ui" sin 2«j PD DC X, = / 2/ _ _ t?i* s in" ari Xo^ ~ ^o ^ 2ui' cos" «i aJo The parameter of the parabola is then - — Vi cos' ari / The di- rectrix is parallel to PD at a distance above the vertex C equal to one half the parameter or — — -- — - , or at a distance of -^ above P. That is, distance of the directrix above P is the height due to the velocity Vi. If we had taken the origin at the vertex C and let Xc and yc be the new co-ordinates, then the horizontal velocity at C would be ■CHAP. III.] TBANSLATIOU — PROJECTILES. 119 Vi COS a, and the horizontal distance passed over in any time t would be Xc = Vit cos ai. The mean vertical velocity would be -ft and 4i the vertical distance yc — -ft^ obtain Combining and eliminating f , we J 2v^ cos" ai aJc' = ^ Vc Avhich is the equation of a parabola referred to its diameter CD and the tangent at the vertex C. The parameter is as before — — . To find the velocity at any point of the trajectory. — The magnitude bf the velocity at any point M is the resultant of the vertical and liorizontal velocities, or, from (1) and (3), i v' = Vx' + Vy' = Vi^ — 2vift sin a, +f^t' (6) The same restdt is obtained at once from the hodograph from the triangle 0pm. Inserting the. value of y from (4), v^ = Vi" - 2fy (7) If the acceleration is due to gravity, we replace / by gr, and have — - = y. But we have just seen that -— is the distance of the 2g 2g ^ '' 2g directrix above P. Therefore -^ y is the distance of the direc- 2g if trix above any point M, and — is the height due to the velocity v. Hence, the speed at any point is the same as that acquired by a body falling from the directrix to that point. To find the direction of the velocity v at any point M, the mag- nitude of which is given by (6) and (7), let a be the angle which it makes with the horizontal. Then we have directly from the hodo- .graph, since angle mOb = a, ■ft; vsma = Vi smai 1} cos a = Vi cos ax. Therefore, from (2), tan a = tan «i — or tan a = tan ai — ft _ ft' = tan ai — Vi cos ai X fx Ui COS-ai (8) (9) To find the time of fiight in a horizontal direction, and the horizontal range. — If in (4) we make ^ = 0, we have for the time Th in which the body reaches the line PX, or the time of flight in a horizontal direction, r. = ?^^i^J^ (10^ J?x, f Inserting this value of < in (2), we have for the horizontal range Bh = 2vx sin ax cos ocx Vx sin 2ax f f (11) >. I f ^ 120 V iT C KINEMATICS OF A POINT — TEANSLATION. [CHAP. Ill, This is twice the distance PD. The greatest value 8in2tri can have is unity, and this occurs when 2«i = 90° or "i = 45°. Therefore the horizontal range, neg- lecting resistance of the air, is greatest for an angle of elevation of 45°, and is equal to -^ . To find the greatest height attained, and the corresponding time. — ut the vertical velocity given by (3) equal to zero, and we have or the time of attaining the greatest height Tv V\ sin ai / (13) just half the whole time of flight as given by (10). Insert the value of the time given by (12) in (4), and we have for le greatest height attained, CD = iJ, H^ Vi sin' ai 2/ (13) Equations (13) and -iJft given by (11) give the co-ordinates of the a vertex C. To find the displacement in any given direction, and the cor- sponding time. — Let 6 be the angle which any displacement FM = makes with the horizontal, then we have BM — y = x tan ^. Sub- tituting this value of y in (5;, we have for the abscissa of the oint M 2ui* cos a, sin (ai — G) _ f i''[8in (2<ari — 0) — sin 6] /cose /cos 6 a^d therefore for the displacement or range Pilf = R, 2vi^ cos ai sin (aj — S) _ t^i^Lsin {2ai — 6) — sin ( R=: (14) (15) /cos'e /cos^e If in (15) we make G = 0, we have the horizontal range Rh as igiven by (11). If wie divide (14) by the horizontal component of the velocity, Vi cosai, we have .1 J.' ^^- I.* 2uisin(ai— 6) ,.„. t = time of flight = :^— (16) ' /cosS This reduces to (10) for 9 = 0. To find the angle of elevation which gives the greatest range in any given direction. — The range R given by (15) is a maximum when sin (2a 1 — 6) is a maximum, or when2ai — 9 = 90° or a j = - (90° + oV 2 V / The direction of projection for the greatest range makes therefore with the vertical an angle 90 — ctri = -- 1 90 — 9 j, that is, it bisects the gle between the vertical and the range. •To find the elevation necessary to hit a given point. — To deter- mine the direction of the velocity Vi in order that the path may pass through a given point given by x and y, we substitute for — 5 — the equivalent value 1 + tan' (n in equation (5), and obtain cos^ ari ^ t once ] tan ai = J- ± 4/ M^ — fx y \fx) 1 + 2u,V (17) fyv^ \ . ~-2_^ ^ —■ - .\ -7_ (l^ ^. 7i rfi, 4u^ IJ^A" tlja^i^ iC^ ^ ^^ CHAP. III.] TRANSLATION — PBOJECTILES. Also from (14) we have «-■ = ^ + 2 sin - ^\ ^ ^, + sm ej, 2 + or, since R cos B = x, (18) 9.1. JfR COS' 6 . . -\ 2 + 2 ^^^"T"!^^'" '"'''• We see from (17) that a, has two values. If a/ is an angle such that sin (2ai' — S) = sin i2ai ~ b), then 2a/ — & = 180" — (2(i-i - 0) or ai' = 90 — (ai — e) and either a/ or ni will satisfy equation (14). With a given acceleration and initial velocity of projection of given magnitude, there are therefore two directions of the initial velocity, ai and 90 — (aj — Oi), and therefore two paths hy which the hody may attain the same point. If in (17) we put (?^)'= 1 + ^^. we have vi" =f{y+ Vx' + 2/" ) and tan a, = -^r-« Smaller values of Vi make tan a, imaginary. Larger values of Vi give two values for tan o-i. In the first case the point cannot be attained. In the second case it would be attained either in the rise or fall of the projectile. To find the envelope of all the trajectories corresponding to different values of ai for a given initial speed Vi. — Equation (5) gives the equation of the trajectory corresponding to the angle of elevation ai. If we substitute 1 + tan" ai for ., , equation (5) becomes y = x tan ai — COS' ai fx\l + tan' ai) 2vi' (19) where x and y are the co-ordinates of any point of the path. For another angle of elevation a/, and the same initial speed Vi, we have , /xi'(l+tan»ai') 2/, = xi tan a/ - ^ 2^ , where Xi and yi are the co-ordinates of any point of the new trajec- tory. If we make x = Xi and y = yi, we have for the point of intersec- tion of the two trajectories, by equating these two equations, ^(tan ai + tan a.O = 1- IT Cro ^ 122 {0 KINEMATICS OF A POINT — TRANSLATION. [CHAP. III. If the angles a/ and ai preaches the limit approach equality, this expression ap- =^tana:i =1, or tan ai = (20) Equation (20) gives then the value of tan a, when the two tra- jectories starting from the same point Pwith the same speed Vi have angles of elevation at P whose difference is indefinitely small. Substituting this value of tan ai in (19) we obtain (21) Equation (21) is then the equation of a curve which passes through all the points in which every two trajectories starting from the same point P at angles of elevation whose difference is indefinitely small cut each other. It is therefore the equation of the envelope or curve which touches all the trajectories or parab- olas described from the same point P with the same initial speed Vu Equation (21) is the equation of a parabola AC A', whose axis PC is vertical, whose focus is the point P of projection, and whose vertex C is in the common direction of the trajectories. With the given initial speed Vi, the projectile can reach any point within this envelope by two angles of elevation and two tra- jectories, as proved page 121. It can reach any point in the en- velope by only one elevation and path. It cannot reach with any elevation and with the given velocity Vi any point outside this envelope. The point, therefore, where this envelope cuts the plane of any given range gives the maximum range in that direction for any given Vu Thus the maximum range on a horizontal plane is found from Vi' (21), by making ?/ = 0, to be -^. The same result is given by (15) when we make ai = 45° and e = 0. Questions of maximum range may thus be readily solved by the equation for the envelope. From (2) we have cos a = —r, and from (4) sin ai = - — ^ — Vit Vit Since cos" ai + sin' ai = 1, we have x:' + {y + ^ff] =vH-' (22) This is the equation of a circle whose radius is Vit and whose centre is situated vertically below P at a distance PD = —ff. The circumference of this circle is reached in the same time by a point starting from P with the velocity Vi in any direction. 'Application of the Calculus.] The same results are obtained by the applica- tion of the differential equations of motion, page 81. Thus in the present case we have for the horizontal component of the Acceleration, since /is vertical, »^'' "" CHAP. III.] TRANSLATIOK— PROJECTILES. 123 and for the vertical component iPy W = -^- (») Integrating (a), since f or « = 0. ^ = v, cos a„ we have '^-/u^ t^o ,^^:y^e. dx - *'■ ^'' '^ ^^''^> ^^=at='"''^^''' (1) Integrating again, since for ^ = 0, a; = 0, we have X = Vit cos OTi (2) Integrating (b), we have, since when t = 0, :~^ = Vi an a^ ~^ ' * Z"'^"' <^ w '^ at J 1 ^'*- -^-v --/T•'-Cw ^y=^ = '"^^^^-ft. . . ^ . '. ..% Integrating again, since ior t = 0, y = 0, we have y = Vit sin ari - -^« (4) Combining (2) and (4) by eliminating t, we have for the equation of the trajectory We have also or, from (1) and (2), «2 = •p," — 2ftvi sin OTi +/*«« (6) Inserting the value of y from (4), «« = tj,2 - 2/y (7) If we differentiate (5), we have, for the tangent of the angle which the velocity at any point makes with the horizontal, dy ^ fx tana = - = tana,-^^,^^ (8) These are the same equations as already given, and from them aU the others are deduced. EXAMPLES. g = 32.16 ft.-per-sec. per sec. Resistance of air neglected. (1) Show that for parabolic motion the hodograph is a straight line. (2) Hie sights of a gun are set so that the ball may strike a given object. Show that when the sights are directed to any other object in the same vertical line, the ball will also strike it. (3) Two bodies projected from the same point in directions mak- ing angles /3, (i' with the vertical pass through the same point in 124 KINEMATICS OF A POINT — TRANSLATION. [CHAP. III. the horizontal plane through the point of projection. If t and t are the times of flight, show that sin {fS - /?0 ^ f' - f sinifJ + /f) ~ r + f """^ifWith what velocity must a projectile be fired at an elevation of 30° so as to strike an object at the distance of 2500 ft. on an ascent of 1 in 40 ? Ans. 311.5 ft. per sec. (5) Find the direction and magnitude of the velocity of projection in order that a projectile may reach its maximum height at a point whose horizontal and vertical distances from the starting-point are b and h respectively. Ans. tan Oi = -— , v, = 4/ ^ JT "^ . \JSf^gun is fired horizontally at a height of 144.72 ft. above the surface of a lake and the initial speed of the ball is 1000 ft. per sec. Find (a) after what time, and (b) at ivhat horizontal distance, the ball strikes the lake, neglecting resistance of the air. Ans. (a) 3 sec. ; (6) 3000 ft. (7) In the parabola described by a projectile, its speed at any point is that tvhich it would have had had it fallen to that point from the directrix. (8) A particle projected at a given elevation with an initial speed Vi reaches the top of a tower hft. high and 2hft. from the point of projection in t seconds. Find (a) the initial speed of another par- ticle which, being projected at the same elevation from a point dis- tant 4h ft. from the tower, will also reach its summit, and (b) the time it will require. Ans. (a) -^^^^_.,(,,J^m±K\ Vh-\-gt' Vg ^-"^j A ball is projected with a velocity of 100 ft. per sec. inclined 75° to the horizon. Find (a) the range on a horizontal plane ; (b) the range on a plane inclined 30° to the horizon ; (c) lohat other direc- tions of the initial velocity would give the same ranges. Ans. (a) 155.5 ft.; {b) 207.3 (V^^ - 1) ; {e) 15' and 45°. (10) Show that with a given initial speed the greatest range on a horizontal plane is just half as great as the greatest range doivn an incline of 30°. (11) Show that if tivo particles meet which have been projected with the same initial speed, in tlie same vertical plane, at the same instant, from tioo given points, the sum of their elevations must b^ constant. (12) On a small planet a stone projected with a speed of 50 ft. per sec. is found to have a maximum range on a horizontal plane of 400 ft. Find the acceleration of falling bodies at the surface of that planet. Ans. 6 25 ft.-per-sec. per sec. (13) Two stones thrown at the same instant from points 20 yards apart, with initial velocities inclined 60' and 30^, respectively, to the horizon, strike a flag-pole at the same point at the same instant. ^^1' ,/^ CHAP. III.] TRAIfSLATIOlf — PROJECTILES. 125 Show that the initial speeds are asl: VB\ and that the distance of th& pole from the nearer point of projection is 10 yards. ^y\\.4c) At what elevation must a body be projected with a speed of ^^ 310.8 /*. per sec. that it may hit a balloon 500 ft. from the earth's surface and at a distance of 1000 ft. from the point of projection. Aus^^r, or 80° 43'. / Vfo) A body is projected with an initial velocity of 30 ft. per sec. inclined 60° to the horizon. Find the velocity after 20 sec. Ans. 617.3 ft. per sec. inclined 148° 36'. 6 to the direction of the initial velocity. (16) If from a point A bodies are projected at the same moment and in the same vertical plane at different angles of elevation, with the same initial speed, Vi, the locus of all the positions occupied at the end of a given time t is a circle whose radius is vit and whose centre is situated vertically below A at a distance ^gt^. J \, [Jitt) A jet of water rises vyith a velocity of 20 ft. per sec. at an ''^' angle of elevation of QQ\ Find (a) the height due to the velocity; (b) the greatest height of the jet; (c) the horizontal range; (d) the time of reaching the horizontal plane ; (e) the height corresponding to the hoHzontal distance 3 ft. Ans. ^af^MT.; (b) 5.17 ft.; (c) 9.24 ft.; (d) 1.14 sec; (e) 4.52 ft. (18) A jet of water discharged horizontally at a height above a horizontal plane of lift, has a range on the horizontal plane ofH ft. Find the velocity of projection. Ans. 15.92 ft. per sec. (19) Prove that the angular velocity of a projectile about the focus of its path varies inversely as its distance from the focus. (20) Shoiv that the envelope of all the parabolas ivhich correspond to a given velocity of projection is equal to the trajectory for which the direction of projection is horizontal. (21) A particle is projected over a triangle from one end of the horizontal base and, grazing the vertex, falls upon the other end of the base. If /3 and y are the base angles and a the angle of projec- tion, shoiv that tan a = tan /3 + tan y. (22) For the greatest range on an inclined plane through the point of projection the direction of motion on leaving is at right angles tp that on reaching the plane. 1 U***SkA particle is projected horizontally with a speed of 32. 16 ft. \ vi per sec. from a point 128.64 /ee* from the ground. Find the direc- tion of its motion when it has fallen half way to the ground. Ans. Inclination to the horizontal = tan - ^ 2. (24) The greatest range on a honzontal plane of a projectile with a given initial speed being 500 meters, show that the greatest range on a plane inclined 60' to the horizontal is 2 — VS kilometers. (25) A stone is let fall in a railway-carriage travelling at the rate of 30 miles per hour. Find its displacement relative to the road at the end of 0.1 sec. Ans. 4.4029 feet, inclined 2° 5'. 5 to the horizon. 126 KINEMATICS OF A POINT — TRANSLATION. [CHAP. III^ (26) The velocities of a projectile at any two points of its path being given, find the difference of the altitudes above a norizontal plane. Fi* — Vi^ Ans. : , where V\ and Fa are the magnitude of the given velocities. (27) A given inclined plane passes through the point of projection of a projectile which eventually strikes the plane at right angles. Find the range of the projectile on the inclined plane, the velocity of projection being given. Ans. If 6 is the inclination of the plane and Vi the velocity of projection, ■A • 2«.' sine the required range is ^ , „ ■ o /» • (28) A particle begins to slide from rest down an inclined plane AB. At the same instant another particle is projected from A. Find the condition that the particles may meet, and ascertain when and where this occurs. Ans. The second particle must be projected at right angles to the plane. If S is the inclination of the plane and Vi the velocity of projection, the time before meeting is < = ^—r, and the distance of the point of meeting from A ° g cos 6 ... , 2»i sin 6 will be 3-r-. g COS'' (29) Prove that the components of the velocities at the extremities of any chord of the path of a projectile, at right angles to the chords are equal. (30) Swift of foot ivas Hiaicatha ; He coiild shoot an arroiv from him, And run forward with such jleetness. That the arrow fell behind him ! Strong of arm was Hiawatha ; He could shoot ten aiirows upward. Shoot them with such strength and swiftness, That the tenth had left the bow-string Ere the first to earth had fallen. Supposing Hiawatha to shoot an arrow every second, and, when not shooting vertically, to have aimed so that the flight of the arrow might have the longest range, find his speed. Ans. About 99 miles an hour. (31) If any number of bodies are projected from the same point in different directions ivith the same initial speed Vi, show that the foci of the parabolas they ivill describe will lie on the surface of a sphere whose radius is ' • ^ 2g (32) The elevation of a projectile is that for maximum horizontal range. Find the time of reaching a point whose horizontal and ver- tical distances from the point of projection are h and k respectively. Ans-^^/^i^EI). (33) If AB is the range of a projectile on a horizontal plane, and t the time from A to any point P of the trajectory, and f the time from P to B, show that the height of P above AB is ^gtlf. CHAP. III.] PROJECTILES — RESISTING MEDIUM. 127 (34) A projectile is fired from an elevation of 1.8 meters above a horizontal plane with a horizontal velocity of 10 meters per sec. How long before it strikes the plane and what is the range f(g= 9.81 meter s-per-sec. per sec.) Ans. 0.6 sec. ; 6 meters. (35) A projectile is fired at an angle of 30° at a target distant 1200 meters in a horizontal direction, (g = 9.81 meter s-per-sec. per sec.) (a) Find the initial velocity. (b) The tim^ of striking. (c) The highest point o/' the trajectory. (d) The velocity of striking. Ans. (a) 116.6 meters per sec.; (6) about 13 seconds; (c) 173.21 meters; (d) same as the initial velocity. (36) A projectile is fired with an initial velocity of 150 meters per sec. from a point 100 meters below a target which is distant horizon- tally 1525 meters, (g — 9.81 meter s-per-sec. per sec.) (a) Find the angle of elevation. (b) The velocity of striking. (c) The time offiight. Ans. (a) 68° 28' 50' for bomb, 25° 16' for ball; (6) 143.3 meters per sec.; (c) 27.715 sec. for bomb, 11.34 sec. for ball. (37) A projectile is fired at an elevation of 45°, and strikes a tar- get at a horizontal distance of 800 meters aria 70 meters lower, (g = 9.81 meter s-per-sec. per sec.) (a) Find the initial velocity. (6) The final velocity. (e) The angle of striking. Ans. (a) 84.95 meters per sec; (6) 93.6 meters per sec; (c)49° 33'. [Motion of a Projectile in a Besisting Medium.] — In tbe preceding examples the motion is assumed to be in vacuo. It should be borne in mind that the re- sistance of the air completely changes the results of the theoretic formulas. The motion of a projectile, taking into account the resistance of tbe air, i» best given by empirical formulas based upon experiment. If, however, we assume that the magnitude of the acceleration decreases directly with the square of the velocity, we may deduce by means of our gen- eral differential equations, page 81, Chap. VIII, the equation of the trajectory for very small angles of elevation. The same method which we shall use holds good for any other assumption as to tbe law of acceleration. The assumption is not strictly accurate, but will serve to illustrate the method of deduction. Let / be the constant acceleration due to gravity, the value of which is given page 111, and let c be the coefficient of resistance, so that c has the value given page 113. Thus, since the velocity at any point is ® = — , we have the retardation — -1 by assumption. We have then from equation (6), page 81, dt) ' since cos a= — , for the horizontal component of the tangential acceleration ds d?x _ /^*V-?- n\ W -~\dt)di ^^ and, since sin a = —-, for the vertical component of the tangential accelera- ds d^ dp = -/-<^)l <^> 128 KINEMATICS OF A POINT — ^TRANSLATION. [CHAP. III. Dividing (1) by — , we have at ax Integrating this, we have ncrn dt logn ^ = - cs-\-C, (3) where C is a constant of integration. Let Vi cos nrj be the component of the initial velocity Vi, parallel to x, ai being the angle of elevation at the point of dx projection; then when « = 0, — = Vi cos ai, and = logn Vi cos a. There- fore, from ^3), dx _c. ^ = e ®i cos tti, (4) where e is the base of the Naperian system of logarithms, 2,718282. Hence dy dy dx -cs dy -jT = ^- Tr = ^ «i cos «! -^ (5) dt dx dt dx If we multiply (2) by dx, and (1) by dy, and subtract, we have ^y^--f-^y = _jax (6) Inserting the value of dt^ from (4), we have d?ydx - d^xdy _ _ fe^ ^^ ^ ^. dX^ Vi^ COS'^ OTi dv The first member of (7) is equal to d-^. We have also /. . dy^\\ ds ^ da dx^ + dy^ = ds\ or {l+£) =-^. or ^^ = Substituting these values in (7), we have i^ + d^J ^^--^.^cos^a,'^ ^^> If we assume the angle of projection a, as very small, so that the trajectory is very flat, we have approximately in such case ds = dx, and « = a;, and II + ;^) = 1* Therefore (8) becomes af=-^^dx (9) dx Vi^ cos^ ai Integrating (9), since when x = 0, -p- = tan ai , we have ^=tana.-^-/-— (e^-1) (10) dx 2c»i* cos'' r' ^ ' "^ ' Integrating (10), we have, since for a; = 0, j' = 0, « = a!tanai-f ir— 5--^^-^ . „ / , (e^- iV . .(11) * ' 2cvi^ cos** aj 4c'»i* cos* ai\ J ^ ' CHAP. III.] PROJECTILES — RESISTING MEDIUM. 129 Equation (11) is the approximate equation of the path. If we expand the last term in a series,. we have y = «tan a, r- ^ — (^9\ If the terms containing c are omitted, and /= fir, equation (12) is that of a parabola, which is the path of the projectile in vacuo. The ordinate of the actual curve is therefore less at any distance x than tha'c of the parabola for the same distance. From equation (4) we have —r- = e~'^ «, cos «! ; and integrating, since for a; = 0, < = 0, we have cvi cos ait = e"" - 1, (13) which gives the time in terms of the abscissa. From (13) we have X = — logn(ciJi cosait-{-l), .... (14) or, in common logarithms, 2.302585 X = c log(cPiCosa,f-f 1), which gives the abscissa x in terms of the time. "The general problem of the path of a projectile in a uniform resisting medium, where the resistance varies as the square of the velocity, was pro- posed by Keill as a trial of skill to John Bernoulli, by whom the challenge was received Feb. 1718. Keill, trusting to the complexity of the analysis, which had probably deterred Newton from attempting any regular solution of the problem in the second book of the Principia, was in hopes that the exertions of Bernoulli would prove unsuccessful. Bernoulli, however, having expeditiously effected a solution, not only of Keill's problem, but likewise of the more general one where the resistance varies as the nth power of the velocity, expressed a determination not to publish his investigation until he had received intimation that his antagonist had himself been able to solve his own problem. He gave Keill till the following September to exercise his talents, declaring that if he received by that time no satisfactory communication, he should feel himself entitled to question the ability of his adversary. At the request of a common friend, Bernoulli consented to extend the interval to the first of November. It turned out, however, that Keill was unable to obtain a solution. At length Nicholas Bernoulli, Professor of Mathematics at Padua, communicated to John Bernoulli a solution of Keill's problem, which the author afterwards extended to the more general one. Finally, on the 17th of November, information was received by John Bernoulli from Brook Taylor, to the effect that he had ob- tain-d a solution. John Bernoulli published his own analysis, together with that of his nephew Nicholas, in the Acta Erudit. Lips. 1719 mai., p. 216." (Walton, Problems in Theoretical Mechanics.) CHAPTER rv. CURVILINEAR MOTION OF TRANSLATION— CENTRAL ACCELERATION. HARMONIC AND PLANETARY MOTION. Central Acceleration. — If the acceleration of a moving point i» always directed towards a fixed point or centre of acceleration, the acceleration is said to be central. The velocity of the moving point at any instant is the resultant of the velocity at the preceding instant and of the integral accelera- tion during the intervening time. But if the acceleration is always directed towards the centre, or fixed point, its moment with reference to that point is zero. Since the moment of the resultant of any two components about any point is equal to the sum of the moments of the components, and since in this case the moment of one of the components, viz., the acceleration, is zero, it follows that the moment of the velocity about the centre, in the case of central acceleration, is constant. Conversely, if the moment of the velocity of a moving point about any fixed point is constant, the acceleration must always be directed towards that point. If r is the distance of the moving point from the fixed point, p the lever-arm, and go is the angular speed at any instant, we have for the moment of the velocity pv — r'ao = c, equal to twice the areal velocity of the radius vector (page 76). Therefore, in all cases of central acceleration -r'ca is constant, or a the area described by the radius vector in a unit of time is constant. c It follows also that in all cases of central acceleration ca = - , or the angular speed is inversely as the square of the radius vector. Cases of Central Acceleration. — The two most important cases of central acceleration are those of harmonic motion, where the central acceleration is directly proportional to the distance from the centre, and planetary motion, where it is inversely as the square of the distance. When the velocity is in the same straight line as the central acceleration we have in both these cases rectilinear motion. The first is simple rectilinear harmonic motion, the second is rectilinear planetary motion or that of a body at great distances from the earth. Both these cases have been considered in Chaps. I and II. When the velocity is not in the same straight line with the central acceleration we have compound harmonic motion and planetary motion in general. The first is of great importance in the study of sound, light, heat, etc., as well as in ordinary kinetics. The second is the motion of planets about the sun and of satellites about their primaries. 130 CHAP. IV.] TRANSLATION — COMPOUND HARMONIC MOTION. 131 Cases of Harmonic Motion. — We have defined simple harmonic motion, page 103, as the motion of a point moving in any path in such a manner that the tangential component of the acceleration a is directly proportional to the distance, measured along the path» from a fixed point in the path. Such motion may be rectilinear or curvilinear. In the first case it is simple rectilinear, in the second simple curvilinear. If the whole acceleration itself, or/, is central, that is, always directed towards a fixed point not in the path, and is always propor- tional to the distance from this fixed point, the motion is central harmonic, or compound harmonic, so called, because it is the result- ant of two simple rectilinear motions, as will be proved in the next article. Simple rectilinear harmonic motion is also central, because the fixed point is in the path. Any Central Harmonic Motion may be Resolved into Two Simple Rectilinear Harmonic Motions at Right Angles. — Let C be the centre of acceleration, and P the position of the moving point at any instant. Let the velocity u of P make an angle a with the axis of X, and let the motion of P be harmonic so that the ac- a' celeration of P is —rV, where a' is the ac- r celeration at a known distance r', and r is the distance CP. The velocity v may be resolved into V cos a and v sin a in the directions GX a' and CY, and the acceleration may be resolved into —rV cos PC A or r — rCA, and -j-r cos PCB or -j^CB, in the same directions. The component accelerations are therefore directly as the dis- tances CA and CB, and the component velocities are in the direc- tions of CA and CB. The central harmonic motion of P, whatever the direction of the velocity v, is therefore the resultant of two simple harmonic motions in the lines CA and CB at right angles. If then any central harmonic motion is resolved into two com- ponents at right angles, the component motions are rectilinear har- monic. Conversely, the resultant of two rectilinear harmonic motions at right angles is a central harmonic motion. Central harmonic mo- tion is therefore called compound harmonic motion. Composition of the Simple Rectilinear Harmonic Motions in Dif- ferent Lines. — ^Let the point M move in a circle AMA' of radius r = CA = CM with a constant angular velocity go. Then the motion of the projection P in the line AA is simple rectilinear harmonic (page 103). Let the point M, move in the circle CBMi of radius ri = CB ~ Ja CMi , with constant angular veloci- ty oai. Then the motion of the pro- jection Pi in the line CB is simple rectilinear harmonic. Let the angle BCA between the planes of the circles be a. Via. 1. 133 KINEMATICS OF A POINT — TKANSLATION. [CHAP. IV. Let the time count from the instant when M, is at B, so that the epoch of Pi is zero (page 105). At this instant let the epoch of P be €. Then e is the difference of epoch, or, in angular measure, the angle of ilf above or below A at the beginning of the time. In any- time t. Ml will have moved from B through the angle wit measured from CB, and M through the angle cot ± e measured from CA. By the preceding Article we can resolve the harmonic motion of Pi into a simple rectilinear harmonic motion at right angles to CA, -and another along CA. The displacement of Pi from C for any time t is Vi cos (oo,t), and this displacement may be resolved into vi cos a cos (ood) along CA, and Vi sin a cos (osit) perpendicular to CA. The displacement of P from C in the same time t is r cos (cot ± e). If a point undergoes these displacements simultaneously, its re- sultant displacement along CA wUl be X = r cos {oot ± e) + Vi cos a COS (ttJif), .... (1) and perpendicular to CA y = Vi sin acos (ooit) (2) The equation of the curve in which the point moves, referred to rectangular co-ordinates with C for the origin, will then be obtained by combining (1) and (2) so as to eliminate t. Such combination (page 131) gives always a central or compound harmonic motion about C, the radius vector from C passing over equal areas in equal times (page 130). Equations (1) and (2) enable us then to find the curve resulting from the combination of any two simple rectilinear harmonic mo- tions inclined at any angle a. If the component motions are at right angles, a = 90°. If the amplitudes are equal, r = ri. If the periods are equal, go = an, the difference of epoch is constant, and, since the epoch equals the pro- duct of the phase at zero of time by Stt radians (page 105), when the periods are equal the difference of phase is constant. When, then, the periods are equal and e = 0, or the epochs are equal, the phases are also equal at any instant. (For definitions of amplitude, period, epoch and phase, see page 105.) Two Component Simple Rectilinear Harmonic Motions in Differ- ent Lines with the Same Period. — In this case go — go, and e is con- stant, or the difference of epochs is constant and difference of phase at any instant is constant. We have then, from (1) and (2), x = r cos (cot + e) + ri cos a COS (aot), y = r sin a cos {cot). Combining these two equations by eliminating oot, we have (ri" sin" a)x' - 2ri sin a{r cos e + r, cos a)xy + (r" + 2rri cos <? cos a+ ri" cos" «)!/" = rVi" sin" a sin" e. (3) This is the equation of an ellipse referred to its centre and rect- angular axes. Hence if a point has two component simple rectilinear harmonic motions in any directions, of any amplitudes, and any difference of epoch, if the periods of the two components are the same, the result- ant motion of the point will be central harmonic in an ellipse, the centre of acceleration at the centrt:. of the ellipse. The areal velocity of the radius vector about the centre is constant (page 130). Such motion is called elliptic harmonic motion. Elliptic har- CHAP. IV.J TRANSLATIOSr — COMPOUND HARMONIC MOTION. 133 monic motion, then, is compound harmonic motion when the periods of the components are the same. Equation (3) gives all cases of compound harmonic motion for equal periods of the components. It will be instructive to derive from it special cases. (a) Two Component Simple Rectilinear Motions in Different Lines with the Same Period and Phase. — In this case we make in (3) e = 0, and therefore the phases are equal, and we have at once r + ri cos a X — . y. Ti sm a This is the equation of a straight line passing through the centre C. The resultant motion is therefore central harmonic in a straight line, or simple rectilinear hai'monic. If CA and CB are the amplitudes r and ri inclined at the angle cr, the result- ant motion has the amplitude CR, in ^ y a direction and magnitude the diagonal of Fiq. 2. the parallelogram whose adjacent sides are r and vi, inclined at the angle a. Conversely, a simple rectilinear harmonic motion whose ampli- tude is CR may be resolved, by completing the parallelogram, into two others in any two directions, of the same period, epoch and phase. Ti If a = 90°, we have y = —x. Therefore the projection of a simple rectilinear harmonic motion on any straight line is also a simple rectilinear harmonic motion of the same period, epoch and phase. If the component motions are more than two, they may be com- pounded two and two, and therefore any number of component simple rectilinear harmonic motions in any directions, of the same period, epoch and phase, give a single resultant rectilinear har- monic motion of determinate direction and amplitude, which may be resolved into two components in any two directions, of the same period, epoch and phase. (b) Two Component Simple Rectilinear Motions in the Same Line with the Same Period and Different Epochs and Phases. — In this case R we make in (3) a = 0, and obtain at once (r* + 2rri cos e + r^)y^ = 0. But since for a = 0, y = 0, (see Fig. 1,) we have r* + 2rri cos e + ri" = constant. In Fig. 3 the points P and Pi move in the line AA' with sim- ple harmonic motion and the diagonal CR = i^r^^ 2rr, cos e + r.', where e is the constant differ- ence of epoch and phase. _ . Since e is constant and CR is constant, its mclmation to CM or CMi is constant. At any instant the resultant displacement is CPi + CP = CS, and the motion of S is therefore the resultant motion and is simple rectilinear harmonic, with the amplitude CR, the diagonal of the parallelogram on r and r.. The epoch and 134 KINEMATICS OF A POINT — TRANSLATION. [OHAP. IV. phase are intermediate between the epochs and phases of the com- ponents. If the epochs and phases are the same, e = and the ampHtude of the resultant motion is r + ri , or the sum of those of the compo- nents. If the difference of epoch or phase is e = tt radians, the am- plitude is r — ri or the difference of those of the components. By taking CMi and CM of proper lengths we can make MCP and MiCM what we please without changing CR. Therefore any simple rectilinear harmonic motion m.ay be resolved into tivo others in the same line, with any required difference of phase and one of them having any desired epoch, the periods being the same. Three or more component simple rectilinear harmonic motions in the same line and with the same period may be compounded two and two, and the resultant will be rectilinear harmonic with the same period. If the periods are different, the angle MiCM= e will vary and CR will vary. When e = 0, CR will have its maximum value r-\-ri. "When the difference of epoch e is tt radians, CR has its minimum value J — ?'i. The angular velocity of CR is also variable. The di- rection of CR will oscillate back and forth about CM, the maximum 1 r\ inclination being sin — . The resultant motion is therefore not simple rectilinear harmonic, but a more complex motion. It is, as it were, simple harmonic with periodically increasing and decreas- ing amplitude, and periodical acceleration and retardation of phase, or epoch. (c) Two Component Simple Rectilineax Harmonic Motions at Right Angles with the Same Period and Different Phases or Epochs. — The general equation for this case is given by (3). If the directions are at right angles, we have a = 90°. Suppose in addition the am- plitudes equal, so that r = ri, and the difference of epoch e = 90°. We have then, from (3), Since the motion is central harmonic, according to page 130 the areal velocity of the radius vector is constant ; and since the radius is constant, the speed in the circle is constant. We have already seen, page 103, that the projection of the motion of a point moving with uniform speed in a circle, upon a diameter, gives rectilinear harmonic motion. The projection upon two diameters at right angles gives then two component rectilinear harmonic motions of the same period, with a difference of epoch of 90°, or of phase of i, since, when one component has its greatest displacement, the other is at the centre with displacement zero. It follows also that two component simple rectilinear harmonic motions at right angles, with the same period and equal amplitudes, differing in epoch by 90° or in phase by one quarter of a period, will give, as a resultant, uniform motion in a circle whose radius is the common amplitude of the components. If the amplitudes are not equal, but oc and e still 90°, and periods the same, we have, from (3), ri'ar' + r^y^ = rVi", which is the equation of an ellipse referred to its centre and axes. The resultant motion is therefore central harmonic in an ellipse, whose semi-diameters are r and r\, the centre at the centre of the ellipse. CHAP. IV.] TKANSLATION — COMPOUND HARMONIC MOTION. 135 The same result is evidently obtained by projecting the circle in the preceding case upon a plane, so as to obtain the required ampli- tude ri, r remaining unchanged. id) Three or More Component Simple Rectilinear Haxmonic Mo- tions in Different Lines with the Same Period but Different Phases or Epochs. — We have seen from equation (3) that the resultant of two simple rectilinear component harmonic motions in any two direc- tions, of the same period and different epoch or phase, is elliptic harmonic motion. We have also seen from (a) that any simple rectilinear harmonic motion may be resolved into two others of the same period and phase or epoch in any two given directions. Any niimber of given simple rectilinear harmonic motions, then, of the same period and different phases or epochs may each be resolved into its own pair in any two given directions. These pairs constitute a number of simple rectilinear harmonic motions in two given lines, all of the same period and different phases or epochs. According to (6), all in one line may be compounded into one re- sultant, and all in the other line into another resultant, these two resultants having the same period and different phases or epochs. The resultant of these two is, according to equation (3), elliptic har- monic motion. Hence the resultant of any number of component simple rectili- near harmonic motions of the same period, whatever their ampli- tudes, directions, phases or epochs, is elliptic harmonic motion, the centre of the ellipse being then centre of acceleration, and the radius vector describing equal areas in equal times. In special cases this becomes, as we have seen, uniform circiilar motion or simple rectilinear harmonic motion. Since the above holds whatever the inclination of the two result- ants, elliptic harmonic motion may be considered as the resultant of two component simple harmonic motions of the same period and different epochs or phases at right angles. Graphic Representation. — We may exhibit graphically simple or compound rectilinear harmonic motion by a curve in which the abscissas represent intervals of time, and the ordinates the corresponding distance of the moving point from its mean position. . In the case of a single harmonic motion we have (page 16£) jc = r cos {oot ± e). If the distance x is to be zero when * = 0, we must have the epoch e =~ radians, or one fourth of the periodic time. This gives a; = r sin oot. Since <» = ^, where T is the periodic time, we have for * = 0, X = 0; for t=^\T,x = r; for f = iT, a; = 0; for t = iT,x = -r; for t=T,x = 0. The curve is the curve of sines, or the curve which would be de- scribed by the point P (page 103) if, whUe M maintained its uniforna circular motion, the circle itself were to move with uniform speed in a direction perpendicular to CA. 136 KINEMATICS 0^ A POINT — TRANSLATION. [CHAP. IV. It is the simplest possible form assumed by a vibrating stringy when it is assumed that at each instant the motion of every particle of the string is simple harmonic. If the rectilinear harmonic motion is compound, we have (page 132) in general X = r cos {cot ± e) -\- Vi cos {ooit ± ei). If the displacement of one of the motions is zero when ^ = 0, we have e = „-; if ei = e + nn, we have X = r sin oot + Vi sin {wit + nit). If the period of one motion is twice that of the other, for instance, we have ooi = 2(», and X = r sin {oot) + Vi sin {2Got + nn). If the difference of phase is zero, n = 0; and if the amplitudes are equal also, we have x = r sin cot + r sin {2Got). This gives a curve as shown in the figure. ' Periods Unequal. — We have in general x = r cos {oot + e), y = ri cos {ooit + ei) for the two component rectilinear harmonic motions at right angles. The elimination of t in any case gives the curve of resultant com- pound harmonic motion. If the periods of the components are as 1 to 2, and e is the difference of the epochs, we have for equal amplitudes X = r cos {2oot + e), y = r cos oot. Eliminating t. ^ = ^|(?^-l)cose + 2|^/l-^.sine[, which is the general equation of the curve for any value of e. Thus for e = 0, or equal epochs, or 2/ (x + A which is the equation of a parabola. For e 2' = 2^i/l y" or r'aP^ — 4y''{r' — y\ which is also the equation of a parabola. CHAP. IV.] TKANSLATIOX — COMPOUND HARMOXIC MOTIOif. 137 If we make ein succession, 0, 1, 2, etc., eighths of a circumference, we obtain a series of curves as shown. Period. 1^2 In the same way we can find the curve for any ratio of periods and difference of epoch. Thus if the periods are as 1 to 3 or 2 to 3, and we make e in succession 0, 1, 2, etc., eighths of a circiunference, we obtain the following series of curves : £=Jf5r --}i7r e-^TT Period 1 : 3 Period 2:3 Blackburn's Pendulum. — The motion of a pendulum which swings through a small arc is, as we shall see hereafter (page 154), simple harmonic, and the projection of the bob on a horizontal plane moves with simple rectilinear harmonic motion. Curves similar to those just given are therefore traced by Blackburn's pendulum. This consists of two pendulums, CED and EB, arranged so as to swing in two planes at right angles. Any difference of period may be made by ad- justing the lengths of the pendulums, and they may be started with any difference of epoch. If the bob B is made to trace its path on a horizontal plane, we have, approximately, the compound harmonic curve. [Application of the Calculus to Harmonic Motion — Let a' be the known acceleration at the known distance ?■'. Then the acceleration at any other dis- tance is —r, where r is the distance from the centre. r' For the acceleration in the direction of the axis of X we have then (f a; _ a' If " ~?'^' ' ' and in the direction of the axis of Y, (Py _ a the minus sign denoting direction towards the centre. (1) (3> 138 KINEMATICS OF A. POINT — TRANSLATION. [CHAP. IV. Multiply (1) by da; and we have d^x , a' Integrating, ,,„ dx= jzdas. dv r lf^y=-^,a,^ + Const. 2\dt j 2r' ^ Let — - = 0, when x = r, then Const. = — -r^ and dt 2r (^y=^Xr^-x^), or dti/^ = - \dt) r y r dx \/r-i — , (3) where we take the minus sign to indicate that x diminishes as t increases, or motion towards the centre. In the same way we obtain froitn (2), if — = 0, when y z=. n, m^:, = --^=-_, (4) where we take the plus sign to indicate that y increases as t increases. Integrating (3) and (4), we have ty%-\-G=cos-^-, or x = r cos<ty% + C Y, f/^+Ci =sin-i^. or y = r,sm\ty^ + Cil, (5) (6) where C and d are constants of integration. Equations (5) and (6) may be written « = r cos C cos < 4/ — — r sin C sin < |/ - = ^1 cos < |/ ^ -j- ^a sin < |/^» y = ri cos Cisinti/ — -\- Ti sin d cos t y — = Bi sin ty % + Bacost y—, where Ai = r cos C, A-t = — r sin C, Bi = Vi cos Ci, Bi = ri sin Ci. If we find from these equations the values of sin and cos in terms of x and y and add their squares, we have, by eliminating t, x\B:' + ^,*) + 2/^(^.2 + A^^) - 2xy{A,Ba + A^B,) = {A,B, - A^B^f . (7) This is the equation of an ellipse referred to its centre and rectangular axes. If we take one of the principal axes corresponding with the axis of X, and dx count the time from the end of this axis, we have for t = 0, y = and — - = and X = r. Therefore, from (5) and (6), (7=0 and Ci = 0, and therefore Ai = 0, Ai = r, Bi = 0, Bi = Vi, and (7) becomes or the equation of an ellipse referred to its principal axes. We have also, from (5) and (6), X = r cos y = Ti sm /a' , dx /a' . /a' CHAP. IV.] TKANSLATION— PLANETARY MOTION. 139 Therefore elliptic harmonic motion can be considered as the resultant of two simple rectilinear harmonic motions at right angles of the same period and different amplitudes, so related that the velocity of one is zero when the ve- locity of the other is a maximum, i.e., one is at the centre when the other is at its greatest range. They therefore differ in epoch by 90°. The time of a complete oscillation is for each of these component motions 2n y --, and this therefore is the periodic time in the ellipse. Planeiary Motion.— Velocity Inclined to "the Central Accelera- tion—Acceleration Inversely as the Square of the Radius Vector— Hodograph a Circle. — Since the acceleration is central, we have (page 130) r'fij = c, or co — -^, where c is a constant and eqticd to twice the areal velocity of the radius vector. Also by assumption we have where a' is the acceleration at a known distance /. Let P be a point which has the ve- ^v locity V, and central acceleration direct- ^ — ed always towards the point O, the ^,„^<^^^^^ radius vector being OP = r. ^"''^V Take O' as the pole of the hodograph y^ V (page 52), and draw OQ parallel and ^ /^ A___b' eqvial to v. Then the tangent to the / o hodograph at Q is the direction of the ' acceleration / at P and is parallel to ^ OP = r. V "^ Since the angular velocity a? at P is /^ ^^^^ the angular velocity of the radius vec- / o'^-^^/^ I \\ tor ?', the angular velocity of the tan- aI ^^^ — njbV gent at Q is also oo. \ ^ j Let C be the centre of curvature of V / the hodograph, so that CQ is perpendic- X^ ^ ^ ^^ ular to the tangent at Q and CQ = p = radius of curvature. Then since the linear acceleration f ot P is f the linear velocity of Q, we have f = poo, or p = —. c a! r'^ ct'/^^ But 00= - and / = — ^ . Hence p = . The radius of r^ r^ c curvature p is therefore constant and the hodograph for planetary motion is a circle. The path which, in consequence of aberration, a fixed star seems to de- scribe is the hodograph of the earth's orbit, and is therefore a circle whose plane is parallel to the plane of the ecliptic. The Path for Planetary Motion is a Conic Section. — Draw OR perpendicular to CQ and therefore parallel to r. OP is the com- ponent of the velocity v in the direction of the radius vector. Draw QN perpendicular to OCB. Then^iV is the component of v per- pendicular to the fixed line CB. But by similar triangles OR QN OR O'C . . ac = CQ^ ^" :p^ = W^'^^''^'''^''*' 140 KINEMATICS OF A POINT — TEANSLATION. [CHAP. IV. that is, the ratio of the velocity along the radius vector to th& velocity at right angles to any fixed line parallel to (yCB is con- stant and equal to e. If then Vi and r-i are the initial and final values of r for an in- definitely short time, and di , d^ the corresponding distances of P from any given fixed line A'B' parallel to O'CB, we have -QN^d7^^.='^ ^^ n-r. = e((i.-d.). . . . (1) Since this holds whenever we take the fixed line A'B\ let u» take the initial distance di such that di = - , or e = — -. Then, from (1), d^ = — , or e = -^, and we have e di II. — Il_ — d\ di that is, the ratio of the distance r of the moving point P from a fixed point O to its distance d from a fixed line has a constant value. This is the Property of a Point Moving in a Conic Section. If e = 1, then O'R = QN. and the pole O' is on the circumference of the hodograph, and the path of P is a parabola. If e is less than unity, the pole O' is inside the hodograph and the path of P is an ellipse. If e is greater than unity, the pole O' is outside the hodograph and the path of P is an hyperbola. When, therefore, a point has a central acceleration inversely proportional to the square of the distance from the centre, it must move in a conic sectio7i with the centre of acceleration as a focus. Conversely, if the path be a conic section and the acceleration is directed towards either focus, it must vary inversely as the square of the distance from the focus. In both cases the radius vector describes equal areas in equal times (page 130). Kepler^s Laws. — By laborious comparison of the observations which Tycho Brahe had made through many years of the planets, especially of Mars, Kepler discovered the three laws of planetary motion which are known as Kepler's Laws. He gave these laws simply as the expression of facts which seemed warranted by the observations. The three laws are as follows : I. The planets describe ellipses, the sun occupying one of the foci. II. The radius vector of each plnnet describes equal areas in equal times. III. The "Harmonic Law," so called. The squares of the period.^ of the planets are proportional to the cubes of their mean distances, from the sun. The second law, as we have seen (page 130), is a necessary con- sequence of central acceleration. From the first law, as we have just seen, it follows that the acceleration must be inversely as the square of the distance. The third law is a direct consequence of such central accelera- tion, as we shall see in the next Article. Verification by Application to the Moon. — Assuming Kepler's third law, Newton was led directly to the conclusion that the ao- CHAP. IV.] TRAIfSLATION — PLANETARY MOTION. 141 •celeration must be inversely as the square of the distance, as follows : The moon and other satellites move around their primaries in sensibly circular orbits, the centre being at the centre of the primary. If T and Ti are the periodic times of two satellites, then accord- ing to Kepler's third law, if r and r. are the radii of the orbits, we must have If 00 is the angular velocity of one satellite, we have (page 76) rGo = -=-, or Gj = — . We have also for the acceleration (page 77) / = - = raj* = -— -. For the other satellite we have in like manner /i = -?jrr- ^® have then /. ^ rv, _ r' / TiV ~ ri"' or the acceleration is inversely as the square of the distance. Con- versely, if the acceleration is inversely as the square of the distance, Kepler's third law is a necessary consequence. The numerical verification of this conclusion by the moon is given in Example 17 (page 55). [Application of the Calculus to Planetary Motion.] — The general formulas for central acceleration have been already given, Chap. VIII, page 85. For any given law of central acceleration we have only to insert the cor- responding value oijfdr in these general equations. (a) To Determine the Path when the Central Acceleration 'Varies Inversely as the Sqnare of the Distance from the Pole. — When the acceleration is inversely as the square of the distance we have where a' is the acceleration at a known distance r', and therefore aV* is con- stant. We have in this case //*=/ a'r'^dr _ a'r'' ~ r Substituting this in equation (42), page 87, we have for the differential equa- tion of the path / <fr* . 1\ , 2a' r'^ ov, taking clB as always positive, dr dS „ A. 2aV^ 1 C If we put z=- — a'r"^, and »* = a'^r* + Cic^ this becomes d9 = , 142 KINEMATICS OF A POINT — TEANSLATION". [CHAP. IV. Integrating, we have 9 = cos'^ — \- Const, n W nen z = n and tnerefore — = ^ — = — j- + 4/ — , let 9 = 0. Then Const. = (p and _i z Q — d> — cos — , or s = «, cos (S — d)). Replacing the values of s and n, we have for the polar equation of the path (1) l + |/l+^cos(e-0) This is the polar equation of a conic section with the pole at a focus. It < will be an ellipse, parabola or hyperbola according as Ci = 0. fdr = , if Ti and Vi are the initial distance and velocity at any instant, Ci = Vi^ , (2) < and according as this is = we have an ellipse, parabola or hyperbola. We > see then that the form of the orbit depends solely upon the magnitude of the initial velocity and not upon its direction. Also if a'r'^ is negative, that is, when the acceleration is directed away from the pole, we have always an hyperbola. From page 100 we have seen that the speed attained by a body starting from rest at an infinite distance from the centre and moving in a straight line with /2a'r'^ an acceleration inversely as the square of the distance is 4/ . ^ ri Hence the orbit will be an ellipse, parabola or hyperbola according as the velocity of projection is less than, equal to or greater than that acquired from an infinite distance. If ei is the angle of ®i with ri , we have from equation (30), page 85, c = ri«i sin ei (3) The constants are therefore given by (2) and (3) and the orbit is determined when the initial velocity «i at the distance ri and with the direction 6i are given. {b) Central Acceleration Inversely as the Square of the Distance from the FocuB — Path an Ellipse. — When the path is an ellipse we have the case of plan- etary motion. Let the point Pmove in the ellipse ABA' with central acceleration always P directed towards the sun 8 in the focus, and B^-'^^ IP"^A varying inversely as the square of the dis- tance or radius vector 8P = r. ' X Let the semi-major axis OA = A and the semi-conj ugate axis OB = B and the eccen- . ■ : 08 tncity -J- = e. Let the angle ASX of the major axis •with any fixed line SX through the focus be 0. CHAP. IV.] TRANSLATION — PLANETARY MOTION. 143 The extremity A of the major axis nearest to the focus S is called in general the lower apsis, or, in the case of a planet, the perihelion. The angle ASX is the longitude of the perilielion. The distance SA is the lower apsidal distance, or the perihelion distance. The other extremity A' is the higher apsis, or, in the case of a planet, the aphelion, and 8 A' is the higher apsidal distance, or aphelion distance. The angle PSX, or the angle of the radius vector with the fixed line 8X, we denote by 6. The angle PSA made by the radius vector with the major axis is then Q — <p. This angle is called the true anomaly. The polar equation of an ellipse with reference to the focus 3 as & pole, counting the angle PSA around from the perihelion, is l + ecos(a-0) ^^ But equation (1), page 142, just deduced, is the equation of a conic section, which becomes an ellipse, therefore, when es^l + ^l^ and ^(1 - e«) = -^, or A = -^^. a^r* ar Ci Inserting the values of (2) and (3), we have (2a'r'*\ e- = l-^ :y^. --^ (5) ar -"ri 2fflV2-ri«,*' (6) where Vi is the initial velocity at the distance Ti making the angle ei with ri. The elliptic orbit is thus determined. From (6) we see that the semi-major axis A depends only on the distance Ti and the velocity of projection Vi and isindependentof the direction of projection €i. In whatever direction the body is projected, the major axis will be the same for the same distance and velocity of projection. We have also «'^'-77T^^ (^> A{1 — r) But we know ,'page 85) that c is twice the areal velocity of the radius vector. If T is the periodic time, then, since twice the area of an ellipse is 27CA^ 4/1 — 6*, we have cT = 2itAW^-e\ or c2 , ,- 47r2^» Aa-e')-"" ~ T' ' But by KBPliER's third law we have for two different planets 'p _ A^ A^ _Ai^ T? ^ A?' ^^ T* ~ Ti^' Hence , or a'r" is constant for aU the planets. ^(1— e**) But «' is the acceleration at a distance r', and aV^ is equal in magnitude to the acceleration at the distance unity, since the acceleration at any distance r . „ «V^ The direct consequence of Kepler's third law, therefore, is that /or aU the planeU the acceleration is tM same at the same distance from the sun. 144 KINEMATICS OF A POINT — TRANSLATION. [CHAP. IV. •' Of all the laws," says Sir John Herschel, " to which induction from pure observation has ever conducted man, this third law of Kepler may justly be regarded as the most remarkable, and the most pregnant with important conse- quences. When we contemplate the constituents of the planetary system from the point of view which this relation affords us, it is no longer mere analogy which strikes us, no longer a general resemblance among them as individuals independent of each other, and circulating about the sun, each according to its own peculiar nature, and connected with it by its own peculiar tie. The re- semblance is now perceived to be a trxie family likeness ; they are bound up in one chain ; interwoven in one web of mutual relation and harmonious agree- ment ; subjected to one pervadmg influence, which extends from the centre to the farthest limit of that great system, of which all of them, the earth included, must henceforth be regarded as members." * (c) Value of a' for Planetary Motion. — In all our equations for central acceleration we see that it is necessary to know the accel- eration a' at some known distance r'. We are dealing in this portion of our subject with Kinematics, or the study of motion only, apart from its causes or the properties of matter. It is therefore not the place here to deal with ideas of " force " and " mass." It is sufficient to state here that the "mass" of a body is the number of standard pounds it will balance at any point of the earth's surface in an equal-armed balance. The unit of mass is then the pound. It will be proved hereafter (see Vol. 2, Statics) that if M is the mass of the sun, and m the mass of a planet, the value of a' which must be used in all equations for planetary motion is given by a = . 9, (1) m where m' is the mass of the earth, and g the mean acceleration of a body at the earth's surface due to gravity. If the two attracting bodies are the earth and a small body of mass m, then a' = — —g, or, since m is insignificant with respect to m', a' = g. If in the preceding Article we had used the value of a' given by <1), we should have obtained M+m ,, 47C^A* , M+m, ,„ 4;r»^i' -m^9r^ = -f^ and —-^gr^=-^, and hence A' Ai' M+m ' M + m,i' We see then that Kepler's third law is not, strictly speaking, ex- a'v'^ act. The value of — ^ , or the acceleration at the same distance, is not, strictly speaking, constant for all the planets. The more accu- rate expression of the third law is that the squares of the periodic times are directly as the cubes of the semi-major axes and inverse- ly as the sum of the mnsses of the sun and planet. The error from this source is, however, too slight to be percept- ible, the mass of Jupiter, the largest of the planets, being less than a thousandth part of that of the sun. * Outlines of Astronomy. CHAP. IV.] TKANSLATION — PLANETARY MOTION. 145 The motion of translation of the planets is not affected by their rotation on their axes, and we may treat them, therefore, as mate- rial points so far as translation is concerned. The sun is not, strictly speaking, a fixed point in this sense, but both sun and planet move in orbits, so that the pole or focus is not at the sun's centre, and this affects the accuracy of Kepler's first two laws. The sun is also attracted by the other planets, and the plan- ets attract each other. The attraction of the planets for each other sensibly modifies their orbits. The ellipse is therefore only an approximation to the path, and requires correction. Kepler's laws are thus only approximate expressions. If there were but two bodies, one fixed and the other free to move, then Kepler's first two laws would be accurate, and the third law would approach accuracy as the mass of the moving body becomes in- significant with respect to the mass of the other. (d) To Find the Velocity of a Planet at any Point of its Orbit. — ^From equa- tion (37), page 87, Chap. VIII, we have = Ci iffdv. fdv = , where a' is given in the preceding Article, and r' is the mean radius of the earth. Also from equation <2), page 142, , 2aV* where Vi is the velocity at the distance ri. Therefore v^=v.^ + 2a'r'^(^^-l^. (1) or, since (page 143) a'r'^ = ■,.■ _ j. , „ o , 2c« /I 1\ From page 85, 'o' = — j, ai^d for an ellipse, from Analytical Geometry, n^ — — -e-j —. Hence ^ 2A - r '• = ffii§ = "^— ) <^ Equation (2) gives the velocity for any distance r if the semi-major axis A is known. Equation (1) becomes the same as equation (2), page 99, for rectilinear mo- tion. COR. 1. We see that the velocity is greatest where r is least, or at pen- he! ion, and least at aphelion, where r is greatest. Cor. 2. If a point moves in a circle of radius r with a speed Vi, its central acceleration is — (page 53). If this acceleration is equal at any mstant to r the acceleration of the planet, we have from equation (7), page 143, Vi^ _ a'r'^ _ ^' •'• V " r^ "" ^(1 — e«) r^' 146 KINEMATICS OF A POINT — TRANSLATION. [CHAP. IV., Therefore, from (2), . c\2A-r) . c^ 1 " A\l -e> ■ A{1 -e'')r' or 9^ : Vi'' :: 2A - r : A. That is, tTie squa/re of the speed in the ellipse is to the squa/re of the speed in the circle as the distance of the planet from the unoccupied focus is to the semi- major axis. Cor. 3. If rj is the perihelion distance and Ti the aphelion distance, we have, from (2), for r = ri , «* = — ; A ri a'r^ r, for r = ra, v^ = — ^ ; A Ta ■while for r = Awe have That is, the speed at tTie extremity of the minor aads is a mean proportional between th^ speeds at perihelion and aphelton. {e) To Find the Time of Describing any Portion of a Planet's Orbit. — From. /a'r''^ fdr = , W^r^ r ""'• From page 143 we have c^ = a'r'^A{l — e^) and Ci = —. A r^A Substituting these values and multiplying by —r-^, we obtain Ar^dr^ Hence at = M \ i rdr ^^^^ [aVy |/4V -{A- rf In order to integrate this expression, let A — r = Aez, then _ / ^Mi (l - ez)dz _ I ^U j - <fe ■ ezdz \ ^*- \a'r^) |/nr^ "WW ( Vi^^ VT^^ r Integrating, t = f-^^*] cos-la -e{l - z^)^\+ Const. \a'r'^ J When z = 1, or r = J. — Ae, or the planet is at perihelion, let t = ti. Then Const. = ti = the time at perihelion. Hence the time for any portion of the path from perihelion is t-ti = l^;Y\ cos-ig - e{l - z'')^}, \ar * / (1> A — T ■where z = — ^ — , or r = All — ez). Ae When 2 = — 1, orr = ^ + Ae, the planet is at aphelion, and t — ti is the time of half a revolution, or , ■where T is the periodic time. CHAP. IV.] TRANSLATION — PLAKETART MOTION". 147 We have then or the square of the periodic time varies as the cvbe of the semi-major axis. EXAMPLES. (1) Find the speed and periodic time of a body moving in a circle at a distance from the earth's centre of n times the earths radius, the acceleration being inversely as the square of the distance. -- = (?)*. --(tT (2) A body at a distance ri from the centre of the earth is pro- jected in a direction which makes an angle of 60° with the distance ri , with a speed Vi which is to the speed acquired by falling from an infinite distance as 1 to ^B ; the acceleration varying inversely as the square of the distance. Find the major axis, the eccentricity, the periodic time and the position of the lower apsis. Ans 2A = -Ti , e =-7r, T = . / i/ — - , where 7*' is the radius of the 2 3 4r r g ' earth. The lower apsis is at a distance ^i from the focus. (3) A body revolves about a centre, the acceleration directed towards the centre and varying directly as the distance. To deter- mine the motion. From the general equations of page 86 we can determine the motion. In a' /* a'r^ the present case we have f = —^, and therefore / fdr = -^. Substituting this in equation (42), page 87, we have for the differential equation of the path J dr^ . 1\ a: or, since dB is always positive, dr y c* cV r«* If we put g = ^ - ■^, and v? = ^^^ , we have -de 2 Vn^ - s« Integrating, we have 1 s 9 = — cos-i [_ Const. 3 n When g = n, let 6 = (p. Then Const. = <p and 1 s (e _ 0) — - cos -1 — , or 2 = n cos 2(0 — 0). 2 n 148 KIlfEMATICS OF A POINT — TRANSLATION. [CHAP. IV. Replacing the value of z, we have after reduction, for the polar equation of the path, 2e^ Ci + 2r«!^ *"''-,sin^(O-0) Ci + inc^ This is the polar equation of an ellipse, the pole at the centre of the ellipse, where Ci + 2nc^' B^ ci + 2nc^ ' Ci — 2nc^ ' For the values of c and Ci we have from equations (30) and (37), page 87, Ci = Vi^ -\ 1—, and c = ?'ii)i sin ei , where Vi is the initial velocity of projection at a given instant, Ti the corre- sponding distance from the centre at that instant, and ei the angle of Vi with Ti. The path is therefore fully determined. To find the periodic time, since the area of the ellipse is tcAB, and since c is twice the area described in a unit of time by the radius vector, we have cT=27tAB, or T=^^^^. c Inserting the values of A and B, we have „ 4*c Inserting the value of w^ we have 2Tt /: This is the same result found in page 107 for rectilinear harmonic motion. (4) A particle describes an ellipse under the action of central ac- celeration, directed to the centre of the ellipse. To determine the law of acceleration. The polar equation to the ellipse, centre pole, is from which we have DiflEerentiating, we have and hence dr"^ dr n 1 \ „ . „ CHAP. IV.] TRANSLATION — PLANETARY MOTION. 149 Differentiating again, = [l--J)[^'''-^^'') But from equation (49), page 88, we have cPQ — . Therefore r (Pr . dr^ 2,1,1 ».2 I ,4 2 ~r Now from equation (45), page 88, we have = cVr-+^i- LVl-iU^4--l -4-^"]- r = -z 7, or r 4- ercosQ = ± Ail — e^). 1 -\- e cos e ' ^ ' The law of acceleration is therefore that of the direct distance. (5) A particle describes a conic section under the action of central acceleration directed to one of the foci. To find the law of accelera- tion. The polar equation of either the ellipse or hyperbola, focus pole, is r Differentiating, dr 4" c cos Qdr — er sin BdQ = 0. Differentiating again, d'r -\- e cos 6d^r — e sin 6d9dr — e sin QdOdr — er cos QdQ^ — er sin Qd^Q = 0. But from eq. (49), page 88, d^B = - ?^^. Therefore JO . /I J /ij«o ^r er cos d^r + e cos 6dr^ = er cos SdS^, or dB^ 1 + « cos 9* Substituting this in eq. (45), page 88, we have cVl e c os 9 \—^^ ^ ____£!___ •' ~ r^^l^r r(l -f e cos B)) ~ r* r(l + e cos 9) ~ ± A{1 - e^)r^ The acceleration is therefore inversely as the square of the distance for either ellipse or hyperbola. The polar equation of the parabola is r = :; 7, or r — r cos 9 = 2A. 1 — cos 9 Differentiating, we have dr — dr cos B -\- r sin BdB = 0. Differentiating again, dV — dV cos e 4- (fr sin 9<Z9 + dr sin 9^9 + r cos 9(Z9« + r sin BcPB = 0. But from eq. (49), page 88, d^B = ; therefore ^ „ « « j/^o <^^ *■ cos 9 (f/. _ (fV cos B = — r cos 9a9*, or -— = — d9» 1 - cos 9 150 KINEMATICS OF A POINT — TRANSLATION. [CHAP. IV. Substituting this in eq. (45), page 88, we have J — r\r'^ /•(! — cos B)) ~ r^ r(l — cos _£__ 2l^ The acceleration is therefore inversely as the sqtLare of the distance for the parabola. (6) A particle describes a hyperbolic spiral under the action of central acceleration directed to the pole. To find the law of accelera- tion. The equation of the hyperbolic spiral, centre pole, is vB = A. We have then edr + rdb = 0, Qd^r -f- drdB + drdQ + rO'Q = 0. From eq. (49), page 88, <Pe = - ?^^. Therefore 6^r = 0, or dV = 0. r From eq. (45), page 88, we have then or the acceleration is inversely as the cvbe of the distance. (7) A particle describes the lemniscate of Bernoulli under central acceleration, the centre being the node. To find the law of accelera- tion. The perpendicular from the node on the tangent is ^ = ± x-jr . Hence dp_Sr^ dp _ 12A* dr ~ 2A'' p^dr ~ r^ * We have from eq. (48), page 88, therefore, J - ^1 ' (8) A particle describes a circle under central acceleration di- rected to a point in the circumference. Find the law of acceleration. The polar equation of the circle is r = 2J? cos 0. Therefore dr = —2B sin 6dd, and cPr = — rd9^ - 2B sin 6d^6. But from eq. (49), page 88, _„ 2drd6 4Ji sin 6d6^ r r ^r = -rd6^-'-^^^^^=rdS^-'-^^. r r and ^ - _ §^ d6^~'^ r ' Substituting in eq. (45), page 88, we have 8W J ^ • CHAPTEK V. <:!ONSTRAINED MOTION OF A POINT. SIMPLE PENDULUM. MOTION ON A CYCLOID. MISCELLANEOUS PROBLEMS. Motion on an Inclined Plane — Uniform Acceleration. — Let a point have a uniform acceleration / in the direction AE, and let the point be constrained to move in the straight line f^ AB which makes the angle a with the horizon. The component of the acceleration in the direc- tion of the motion is then / sin a. The motion along AB is then rectilinear motion under uniform acceleration / sin a, and equations (1) to (6), page 93, apply directly, if we substitute / sin a in place of g. If Vi is the initial velocity at A and v is the velocity at B, wo have from (5), page 93, if — vi' = 2/1 sin a, 'where I is the length of the inclined plane AB. But I sin a = AE. The speed, therefore, gained in moving from A to B is equal to iliat which would he gained in falling through AE with the uniform acceleration f. The time in falling from A to ^ is from (1), page 93, f = — ^, and in passing from A to J5, f = V— Vi Hence /sin a 1=1- t AE' or the times are proportional to the distances I and AE. The distance passed through along AB is from (3), page 93, l = Vxt -V —/sin a. <', where V\ is the initial velocity. If the point starts from rest, we have for the distance along AB I = 7:/ sin a . f. 2'' Let AD be the vertical diameter of a circle and AB = I any chord. Join DB. Then we have AB = AD cos DAB = AD sin ABC. If AB = I, we B have also AB = ^ff sin ABC. Therefore AD = \ft^, or t =|/^^. This is independent of the posi- tion of the chord AB, and therefore t is the same for any chord through A or D. • ' 151 152 KINEMATICS OF A POINT — TRANSLATION. [CHAP. V, Hence for uniform acceleration /, the time of descent down alt chords through the highest and lowest points of a circle are equal. This property enables us to find the line of swiftest descent to a given curve from any point in the same vertical plane. Thus if EG is the curve and A the pointy draw AO vertical or parallel to the direc- tion of/, and with centre in AC describe a circle passing through A and tangent to the curve EG at P. Then AP is the line of swiftest descent from A to the curve EG. For any other point p in EG, Ap cuts the circle in some point q, and since the time from A to g is equal to that from A to P, the time from A to- p is greater. EXAMPLES. g = 32.16 ft.-per-sec. per sec. Friction, etc., disregarded. (1) Find the position of a point on the circumference of a vertical circle, in order that the tim£ of rectilinear descent from it to the centre may be the same as the time of descent to the lowest point. Acceleration due to gravity. Ans. 30° from the top. (2) TTie straight line down which a particle will slide, under the action of gravity, in the shortest time from a given point to a given circle in the same vertical plane, is the line joining the point to the upper or lower extremity of the vertical diameter, according as the point is tvithin or without the circle. (3) Find the line of quickest descent from the focus to a parabola whose axis is vertical and vertex upwards, and show that its length is equal to that of the latus rectum. Acceleration vertical and uniform. (4) Find the straight line of swiftest descent from the focus of a parabola to the curve when the axis is horizontal. Acceleration vertical and uniform. (5) The times in which heavy particles slide from rest down in- clined planes of equal height are proportional to their lengths. (6) Show that if a circle be drawn touching a horizontal straight line in a point P and a given cuirve in a point Q below P, PQ is the line of swiftest descent to the curve, under constant vertical accel- eration. (7) Find the straight line of quickest descent from a given point to a given straight line, the point and the line being in the same vertical plane. Acceleration constant and vertical. Ans. From P, the given point, draw a horizontal line meeting the given line in C. Lay off along the given line CD equal to PC. PD is the required line of swiftest descent. (8) A given point P is in the same plane with a given vertical circle and outside it, the highest point Q of the circle being below P. Find the line of slowest descent from P to the circle. Acceleration constant and vertical. Ans. Join PQ and produce it to meet the circumference in B. PR is the line required. CHAP, v.] TRANSLATION — CONSTRAINED MOTION. 153 (9) A number of heavy particles start ivithout velocity from a common point and slide down straight lines in various directions. Show that the locus of the points reached by them, with a given speed is a horizontal plane, and that of the points reached by them in a given time is a sphere whose highest point is the starting-point. (10) The times required by heavy particles to descend in straight lines from the highest point in the circumference of a vertical circle to alt other points in the circumference are the same. Also to descend in straight lines to the lowest point in the cir- cumference, from all other points in the circumference, the times are the same. (11) If heavy particles slide down the sides of a right-angled triangle ivhose hypothenu^e is vertical, they will acquire speeds pro- portional to the sides. (12) A point having a constant acceleration of 24: ft.-per-sec. per sec. is constrained to move in a direction in which its speed changes in 1 minute from 10 to 250 yards per sec. Find the inclination of its direction of motion to that of the given acceleration. Ans. 60°. (13) A heavy particle is projected up an inclined plane whose in- clination to the horizon is §0". Find the distance traversed during a change of speed from 48 to IQft. per sec. Ans. 63.68 ft. (14) A point has, when 1 mile up an incline of 1 in 50 (i.e., one having an inclination to the horizon of sin' ^ ~], an upward ve- locity of 30 miles an hour, (a) In what tim^ will it come to a stand- still f (b) If it afterwards slides down, with what speed will it reach the foot of the incline f Ans. (a) 1 min. 8.4 sec. (6) 63.7 miles per hour. (15) A body slides from rest doivn a smooth inclined plane and then falls to the ground. The length of the plane is IS ft., its incli- nation to the horizon 30°, and the height of its lowest point from, the ground 40 ft. Find the distance horizontally from the end of the plane to the point where the body reaches the ground. [Take g = B2 ft.-per-sec. per sec] Ans. 15 V'S'ft. Motion in a Curved Path— TTniform Acceleration. — Let ABCD be any curved path, and Ad the direction of the acceleration/. Any very small portion of the curve, AB, may be considered as a straight line. We have then, as on page 151, the change of speed in moving from A to B, the same as in moving from A to 6 with the con- stant acceleration /. So also, in moving from B to C, the change of speed is the same as in moving from 6 to c with the constant acceleration /. Hence the change of speed in traversing any portion ot the patn AD is the same as in traversing with constant acceleration / the projection Ad of the path on a line in the direction of the accelera- tion. 154 KINEMATICS OF A POINT — TKANSLATION. [CHAP. V. If then Vi is the initial speed at A, and vis the speed at any point J), we have v'-v:' = 2f.Ad. tion in a Circle — Uniform Acceleration. — This is the case of the simple pendulum, which consists of a heavy- particle attached to a fixed point by a massless in- extensible string. Let C be the point of suspension and CA the radius, and let the acceleration / be uniform and vertical. For any position of the point i^ the angle ACP= &, and the acceleration may be re- solved into a tangential component / sin 6 and into a normal component /cos 9. The normal component has no effect upon the motion in the curve at P. If the angle 6 is very small, the arc will not differ materially from the sine, and we have sin 6 arc AP , where I is the length of the radius CA. The tangential acceleration at the point P is then a = / X arc AP It is therefore directly proportional to the displacement of P from A, measured along the path. The motion of P is thus simple harmonic motion about A as a centre (page 103). The periodic time is then (page 104) ' = 27t\/- displacement tangential acceleration = 2;r ^ 9 or for the simple pendulum the time of a vibration is t The periodic time is therefore for small displacements independ- ent of the amplitude, and therefore for small arcs the oscillations are isochronous. The time of oscillation is usually taken as half the periodic time, or the time between the instants at which the pendulum reaches opposite ends of its swing. Thus the seconds pendulum makes a complete oscillation in 2 seconds. If S is not very small the time is different, but the variation is practically very slight. (See page 160.) CoR. If the velocity of P at any instant is not wholly in the plane PCA, it may be resolved into two components, one in the plane PC A and the other perpendicular to it, and both tangential to a spherical surface Hence, in the case in which is small, P's motion may be resolved into two simple harmonic motions of the same period; and its motion is therefore (page 135) elliptic har- monic motion, the period being the common period of the compo- nents The ellipse described will depend upon the amplitude and epoch of the components and therefore upon the magnitude and direction of the initial velocity of P. If G is not very small, and the component motions are of differ- ent amplitudes, the periods will have different values, and the point P describes curves similar to those given on page 137. If the component motions are equal in amplitude and therefore CHAP, v.] TRANSLATION— CONSTRAINED MOTION. 155 in period and differ in phase by one quarter period, the point P moves (page 134) in a circle about the foot of the perpendicular on CA as a centre. This is the case of the conical pendulum. EXAMPLES. (1) Find the time of oscillation of a pendulum 10 ft. long at a place at which g = S2ft.-per-sec. per sec. Ans. 1.75 sec. (2) Find the length of the seconds pedulum at a place at which fif = 31.9. Ans. 3.232 ft. (3) Find the length of the pendulum which makes 24 beats in 1 min. when g = 32.2. Ans. 20.39 ft. \^i(i)A seconds pendulum is lengthened 1 per cent. How much does it lose per day f Ans. 7 min. 8.8 sec. (5) The length of the seconds pendulum being 99.414 cm., find the valine of g. Ans. 981.17 cm.-per-sec. per sec. (6) A pendulum 37.8 inches long makes 182 beats in 3 min. Find the value of g. Ans. 31.78 it.-per-sec. per sec. \0>ff)lf two pendulums at the same place make 25 and 30 oscilla- tions respectively in 1 sec, what are their relative lengths f Ans. 1.44 to 1. (8) A pendulum which beats seconds at one place is carried to another where it gains 2 sec. per day. Compare the values of g at the two places. An^ As 0.999953 to 1. ^""^(y) A pendulum which beats seconds at the sea-level is carried to /^^ ,.-.«► the top of a mountain where it loses 40.1 sec. per day. ^ 7"TfiiTli(ifty- /^^ the value of g to be inversely proportional to the ^t/tm^^rom the hs cO\a^^^K centre of the earth, and the sea-level tobe 4000 miles from that point, U find the height of the mountain. Ans. 1.86 miles. Motion in a Cycloid — Uniform Acceleration.— A cycloid is the curve traced by a point in the circumference of a circle which rolls allong a straight line. If the circle -EP rolls along the Hne AB, the point P being origi- nally at A, the path of P is the cycloid ACB. 156 KINEMATICS OF A POINT — TRANSLATION. [CHAP. V. If C is the position of P when the diameter of the circle through Pis. perpendicular to AB, the line CD perpendicular to AB is the axis and C is the vertex of the cycloid. Let the uniform acceleration / be always parallel to DC and ver- tical. Let the moving point Q have at ^i a speed zero. Its speed at Qi is then (page 154; v'' = 2f.N,N^. Let t be the time in which the point would with the same ac- celeration and with initial speed zero move from D to C Then GD = \ft\ Hence a u^ = ^ . N.N. . CD = \CD{CN - CN,). t t Now by a property of the cycloid 4CD . CN = CQ.' and 400 . CN^ = CQ^. Hence ^ = \{CQ.- - CQ^')- 2CD Now f = — — is a constant. Hence the motion of Q in the cycloid is simple harmonic (page 103), where — = — , a' being the tangential acceleration of Q at the distance s' measured along the curve. If T is the time of a complete oscillation, we have T=2ic i/% = 2nt = 2n i/^. y a' ^ f It t' is the time occupied in moving from Qi to C, ^2CD </' or the time of a pendulum whose length is 2CD, or 4 times the radius of the generating circle. As this involves only constant quantities, the time is the same whatever be the position of the starting-point ^i, or the oscillations are isochronous. Hence the cycloid is called a tautochrone. This result is rendered of practical importance by one of the properties of the cycloid, viz., that if a flexible and inextensible string AB is fixed at the end A and wrapped tightly round the semi-cycloid AC, the end B of the string as it unwinds will describe another semi- cycloid. If then AC and AD are fixed semi -cycloids, symmetrical with refer- ence to the vertical AB, and AB is a simple pendulum, B will describe a cy- cloid, and its oscillations will be isochronous whatever their extent. CHAP, v.] TRANSLATION — CONSTRAINED MOTION. 157 [Application of the Calculus. — To Determine the Motion of a Point Constrained to Move in a Cycloid, the Acceleration being Constant, in the Direction of the Axis and towards the Vertex.] — By the application of the general formulas of page 88 we can deduce the results A_ d B already obtained. Let the axis CD = 2r, where r is the radius of the generating circle DF'C. Let the acceleration / act downward. Let CN = y, NP— x and the length of arc CP = s. Let the initial position be P, at the height CNi = h above C, and the speed at Pi be «i = 0. We have thus the case of equation (55), page 89, and obtain at once, since / is negative and Si = h, «= 4/a/(A - y) for the speed at any point given by CN — y. When y = 0, we have, at the lowest point G, v = 4/2/%, which is the same as that due to the vertical height h. By the property of the cycloid we have « = arc CP= 2 VDG X ON = 2 )/Wy = 2 chord CP'. Hence ds= ± dyif — ^ y We substitute the minus value in equation (56), page 89, because for de- scent the arc decreases as the time increases. We have then dt -s/% dy f Vhy-y^ Integrating, since for < = 0, y = h, we have <=|/^.(;r-versin-^^). (1) For the time of descent to the lowest point where y = 0, or for the time of one quarter of a complete oscillation, The periodic time is then , /r Tt . /At < = ^r 7 = 2-r 7 ^=2* or the same as a simple pendulum (page 154) whose length is 4 times the radius of the generating circle DP' C. 1, x xi • The time is independent of h and is the same no matter what the posi- tion from which the point begins to descend. The oscillations are therefore isochronous and hence the cvcloid is called the tautochrone. The reason of this remarkable property is easily seen by considering the tangential acceleration. _,„ m. In the cycloid the chord CP' is always parallel to the tangent IP. Ihe tangential acceleration or tangential component of/ is then J9 CP' g=:/sinriy=/sinC7i)P'=/^^- /; 4r The tangential acceleration is therefore directly proportional to the distance from the vertex measured along the path, and the motion of P is simple har- monic (page 103). 158 KINEMATICS OF A POINT — TRANSLATION. [CHAP. T^ The periodic time is then (page 104) / displacement / « i/^ — " Y tangential acceleration ~ \/f* ~~ f' V J^ as already found. A 1 TT /T If in (1) we make y = — = ^GNi, we have t = jr-^/ -;?, or half the time « « a ' J from Pi to C. The time, therefore, in descending through half the vertical space to C is half the time of passing from P^ to G. [To Find a Curve such that a Point Moving on it under the Action of Gravity will Pass from any one Given Position to any Other in Less Time than by any Other Curve through the Same Two Points.] — This is the celebrated problem of the ' ' curve of swiftest descent " first propounded by BemouUi. The following is founded upon his original solution. If the time of descent through the entire curve is a minimum, that through any portion of the curve is a minimum. It is also obvious that between any two contiguous equal values of a con- tinuously varying quantity, a maximum w minimum must lie. This principle though simple is of very great power, and often enables us to solve problems of maxima and minima, such as require not merely the pro- cesses of the Differential Calculus but those of the Calculus of Variations. The present case is a good example. ,p Let, then, PQ, ^i?and PQ;, Q^B be two pairs of indefinitely small sides of a polygon such that the time of descending through either pair, starting from P, may be equal. Let QQ' be horizontal and in- definitely small compared with PQ and QR. The curve of swiftest descent must lie between these paths, and must possess any property which they have in common Hence if we draw Qm, (^n perpendicular to RQ, PQ, and let v be the speed down PQ or PQ^ (supposed uniform) and v' that down QR or QR, we have for the time from P to i2 by either path V "^ tf V ' v' ' V v' * or Qn _ Q'm V ~ if ' Now if 9 be the inclination of PQ to the horizon, and Q* that of QR, we have ^ = Q(^ cos 6, Q'm = Q(^ cos &'. Hence This is true for any two consecutive elements of the required curve, and therefore throughout the curve we have, at any point, v proportional to the cosine of the angle which the tangent to the curve at that point makes with the horizontal. But «' is proportional to the vertical distance h fallen through. Hence the curve required is such that the cosine of the angle it makes with the horizontal line through the point of departure varies as the square root of the distance from that line Now in the figure of page 157 we have, from the property of a cycloid. cos CP'N = cos TPN = cos GDP' = ^^ = y' DP' _ /DN DG ~y DG' The curve required is tJierefoi'e the cycloid. The cycloid has received on account of this property the name of BrachistocJirone. CHAP, v.] TRANSLATION — CONSTRAINED MOTION. 15* [To Determine the Uotion of a Point Constrained to Hove in a Circle, the Ao- celeration being Constant and Vertical.] — This is the case of equation (55), pag» 89. A c B Let DN = y, NP = x. Let the speed at Pi be Di = 0, the distance DNi = h. For the speed at any other point P we have \ ^i at once from equation (55), page 89, since fy is minus and «i = h. V = y2f{h - y), VFhere y is the distance DN. When y = 0, we have for the lowest point, D» which is the same as that due to the vertical height h. From equation (56), page 89, we have dt * The equation of the circle referred to 2) as origin is x^ = 2rx — y^, where r is the radius. Hence 2r2/-y* But cZ«» = dx» + ^'^ = d/(l + ^^^) •. <fe= ± 2ry-y^' rdy 4/2ry — 2/* We substitute the minus value in equation (56), page 89, because the arc- decreases as the time increases, and obtain , r dy ^ ^_ ^y (1^ VW VQi-yt^ry-f) V^f Vihy-y'>){2^-y) If y is small in comparison to 3r, this reduces to 1 ./r^ dy ^y f' 4/M dt ^hy - y* Integrating, since when i = 0,y = h,we have When y = 0, It , pr This is the time of one quarter of a complete oscillation. The periodic time is then '=2;r|/^. This is the same result as on page 154. v. t n\ If, however, y is not small in comparison to 3r, we have, from (1), VW V^y-y'' ^ ^ ^hy-y'^ 160 KINEMATICS OF A POINT — TEANSLATION. [CHAP. V, Expanding [ 1 — 5 1 by the binomial theorem, we have Thus the terms to be integrated are of the form . the exponents yhy - y'^ n being the natural numbers beginning with zero. Performing the integra- tions, and taking the limits y =^ h and ^ = 0, we have for the time of one quarter oscillation This series converges more rapidly as h becomes smaller. If we take only the first term we have, as already found, t = —4/ —. Q •' We can put h = r — r cos 6 = Sr sin^ — -, where 6 is the semi-angle of oscil- lation DGPi ; or taking the arc as equal to its sine, we have for the first two terms = yj(+0 Thus if the point swings through an arc of — radian or 5°. 7, on each side of the vertical, the time of an oscillation is increased by about , or in the case of the seconds pendulum the time of a beat is increased by of a second. For a swing on either side of the vertical of any amount we have for the time of a quarter oscillation . f i /r(, , 1 . , 6 , 9 . ^ 6 , 225 . , 6 \ ' = ■2^/11+4 "'^'y + ei ""' 2 +2304 ""' 2' 'H For a swing of 60° on either side we have, therefore, *- 2r /l^+i6 + 1024J' or the time is increased by about t^. MISCELLANEOUS PROBLEMS. 161 MISCELLANEOUS PROBLEMS. We give here as exercises for the student a number of problems covering the preceding Chapters. Resistance of the air and friction are neglected. (1) A courier travels at the rate of 5^ miles in 8 hours. Six hours after his departure another courier is dispatched from a place 12 miles behind the starting-point of the first. The second courier travels at the rate of 8 miles in 10 hours over the same route as the first. How long before he will overtake the first ? Ans. 120 hours. (2) A man walks on the deck of a vessel from bow to stern at the rate of 3 miles an hour, ivhile the vessel moves at the rate of 7 miles an hour. Find the speed of the man with reference to the earth's surface. Ans. 4 miles an hour, in the direction of the ship's motion. (3) A point moves in a given path for 10 seconds with a uniform rate of change of speed of + 8 jt.-per-sec. per sec. Find the final speed and the space traversed, if the point starts from rest. Ans. v = at, s — -^afi ; « = 80 ft. per sec, s = 400 ft. (4) A point has an initial speed of 7.7 ft. per sec. and a uniform rate of change of speed of + 5.5 ft.-per-sec. per sec. Find the time of passing over 2200 ft. Ans. 8 — Vit + -^af. < = 26.9 sec. 2 (5) A point has an initial speed of 7 ft. per sec. and a final speed of 125 ft. per sec, and describes a distance of 3250 ft. What is the uniform rate of change of speed f Ans. s — Si — — jr , a = 2.4 ft.-per-sec. per sec. (6) A point has an initial speed of 100 ft. per sec. and a rate of change of speed of 1 ft.-per-sec. per sec. Its final speed is 7 ft. per sec. Find the time of motion and the space described. Ans. V = Vi — at, < = 93 sec. ; s — Si = Vit — -^afi, 8 — 8i = 4975.5 ft. (7) A locomotive has a speed of 30 miles an hour on a level when the brakes are applied. The loss of speed due to the brakes is 8 ft.- per-sec. per sec. Find (a) the speed at the end of 3 seconds and the distance traversed ; (b) the time of coming to rest; (c) the retarda- tion in order that the locomotive may come to rest in 30 seconds. Ans. » = «! — a<, ?)i = 44 ft. per sec; o = 8 ft.-per-sec. per sec; a — 8i = Vit — -^at^. (a) 20 ft. per sec, 118 ft.; (5) t = 5.5 sec; (c) 1.47 ft.-per-sec. per sec. 162 KINEMATICS OF A POINT— TKANSLATION". (8) A body falls for 4 seconds in vacuo. Find the final velocityr and the space described (g = 32^ ft.-per-sec. per sec). Ans. v = gt, 8 — ^gi^; V = 138f ft. per sec, s = 257i ft. (9) A body falling in vacvo has a final velocity of 250 ft. per sec. Find the time of falling from rest, and the distance described {g = 32^ ft.-per-sec. per sec). Ans. 7.77 sec; 971.25 ft. (10) A body falling in vaxiuofrom rest describes a distance of 85 feet. Find the tims of fall and the final velocity (g = 32^ ft.-per- sec. per sec). Ans. 2.3 sec. ; 73.9 ft. per sec. (11) A body is projected vertically uptvards in vacuo icith a ve- locity of 40 ft. per sec. Find the height and the time of ascent (g = 32| ft.-per-sec. per sec). Ans. 24.87 ft.; 1.24 sec (12) A body projected vertically upwards in vacuo returned after 18i seconds. Find the initial velocity and the height of ascent {g = 32^ ft.-per-sec per sec). Ans. 297.54 ft. per sec; 1376 ft. (13) A body falling in vacuo has at a given instant a velocity of 17 ft. per sec, at a later instant a velocity of 90 ft. per sec. Find the time between the two instants and the distance traversed (g = 32^ ft.-per-sec per sec). Ans. 2.27 sec; 121.44ft. (14) A stone is dropped into a well and the splash is heard in 3 seconds. If sound travels in air with a uniform velocity of 1090 ft. per sec., find the depth of the well (g = 32^ ft.-per-sec per sec). Ans. 130.4 feet. (15) A point has two component velocities (or accelerations), at right angles, of 35 and 87 units. Find the resultant velocity (or ac- cslerations). Ans. 93.77 units, making an angle with the 35 units of 68° 5'. (16) A point has a velocity of 120 ft. per sec. Resolve this into two component velocities at right angles, (a) one of the components being 75 ft. per sec; (b) one of the components making an angle of 34° 7' 3" with the resultant. Ans. (n) 93.68 ft. per sec, the resultant making an angle with 75 ft. per sec of 51° 19' 4". (b) 99.343 ft. per sec. adjacent to the given angle and 67.306 ft. per sec. opposite. (17) A point has two accelerations of 115 and 89 ft.-per-sec per sec. at an angle of 147° 8' 3 . (a) Find the resultant acceleration ; (b) Find the angle between the given accelerations ichen the resultant is equal to the lesser ; (c) when it is equal to the greater. Ans. (a) 62.865 ft.-per-sec. per sec, making the angle of 88° 4' with 89 ft.- per-sec. per sec; (b) 130° 14' 44"; (c) 112° 45' 54'. (18) A point has an acceleration of 77.5 ft.-per-sec. per sec. Re- solve it into two components (a) making unth the given acceleration the angles 35° 7 11 ' and 52° 9' 8"; (b) ivhen one of the components is 50.5 ft -per-sec. per sec. and makes an angle of 36° 8' 6" with the re- MISCELLANEOUS PROBLEMS. 163 sultant ; (c) when one of the components is 60 ft. -per-sec. per sec. and the other makes an angle of 47' 10' 11" with the resultant; (d) when the two components are 46.2 and 35 ft. -per-sec. per sec. Ans. (a) 61.265 and 44.634 ft. per-sec. per sec. (6) 47.37 ft.-per-sec. per sec, making an angle witli the resultant of 39° 2' 8". (c) 71.88 or 33.48 ft.-per-sec. per sec, making an angle with the resultant of 61° 38' 17" or 34° 9' 4". {d) The components make an angle of 35° 4', and the resultant makes with the component 46.3 the angle 15° 3' 18". (19) A stream flows ivith a velocity of 1 ft. per sec, and a boat whose speed is 1.3 ft. per sec. is steered up stream at an angle of 30° witJi the current. Find the resultant velocity. Ana. 0.66 ft. per sec. down stream at an angle of 79° 3' with the current. (20) What is the ratio of the speed of light to that of a cannon- ball moving at the rate of 2400 feet per sec, if light passes from the sun to the earth, a distance of 91,500,000 miles, in 8^ minutes f Ans. 403600 to 1. (21) ABC is a triangle. Two spheres of radii rx and r, start to- gether from A and B, their centres moving along AC and BC ivith velocities which carry them separately to C in the same time. Find the distances each has gone through when they meet. Ans. If a, b and c are the sides of the triangle, the required distances are -(c — ri — Ti), -(c - ri — r»). (22) If a particle is projected vertically in vacuo with a velocity of 8g, find the time in which it will rise through the height 14gf. Ans. 3 sec. and 14 sec. (33) A body falling in vacux) is observed to describe 144. 9 ft. and 177. 1 ft. in two successive seconds. Find g and the time from the beginning of the motion. Ans. g = 'd2.2 ft.-per-sec per sec ; < = 4 sec. to the beginning of the first of the two seconds. (24) A, B, C, D are points in a vertical line, the distances AB, BC, CD being equal. If a particle fall from A, prove that the times of describing AB, BC, CD are «s _ 1 : \/2- 1:VS- V2~ (25) A particle describes in successive intervals of 4 seconds each spaces of 24 and 64 feet in the same straight line. Find the accelera- tion and the velocity at the beginning of the first interval. Ans. 3.5 ft.-per-sec. per sec; one foot per sec. (26) A particle moves 7 ft. in the first second, and 11 and 17 ft. in the third and sixth seconds respectively. Show that these facts are consistent with the supposition of a uniform acceleration. (27) A falling particle is observed at one portion of its path to pass through nft. in s seconds. Find the distance described in the next s seconds. Ans. ?i-f-^s'^. (28) If s, ms, are the spaces described by a body in times t, nt, respectively, determine the acceleration and the velocity of projection. 2(m — n)s J (m — n^)s Ans. — ; -T-, and —- --. n{n — \)t^ n{l — n)t 164 KINEMATICS OF A POINT — TRANSLATION. (29) If the focus of the path of a projectile he as much below the horizontal plane through the point of projection as the highest point of the path is above it, to find the angle of projection. Ans. If a'j is the angle of projection, Ui = tan -if — ■=- j. (30) Particles are projected from the same point in the same di- rection with different speeds. Find the loctts of the foci of their joaths. Ans. A straight line through the point of projection making an angle with It the horizontal equal to — — 2ai , where ax is the angle of projection. (31) If a particle is projected in a direction inclined to the hori- zon, show that the time of moving between two points at the extrem- ities of a focal chord of the parabolic path is proportional to the product of the velocities of the particle at the two points. (32) Two particles are projected from two given points in the same vertical line in parallel directions and with equal speeds. Prove that tangents drawn to the path of the lower will cut off from the path of the upper arcs described in equal times. (33) If a particle is projected from a given point so as to strike an inclined plane through that point at right angles, prove that tan {a I — B) = —cot a, where ai is the angle which the direction of projection makes with the horizon, and Q is the inclitiation of the plane to the horizon. (34) A particle is projected from a given point with a given velocity. Find the direction of projection in order that its path may touch a given plane. Ans. Let ai be the angle of projection with the horizontal and /i the angle of the given plane with the horizontal, •Pj the velocity of projection and d the distance from the point of projection to where the given olane cuts the hori- zontal through the point of projection. Then cos (90 — /3 — ttj) = -^-^ —. Vi (35) To find the least velocity with which a body can be projected from a given point so as to hit a given mark, and the direction of projection in this case. Ans. Let d be the horizontal distance from the point of projection to the mark, Vi the velocity of projection, ai its inclination to the horizon and /3 the angle of elevation of the mark above the horizon. Then •■=/^"^^^. ..=:-(^+4 (36) If two circles the planes of which are vertical touch each other internally at their highest or lowest points, and if any chord be drawn within the larger circle, terminating respectively at its highest or lowest point, prove that the time of descent down that por- tion of the chord which is exterior to the smaller circle is invariable. (37) AP, PB are chords of a circle, AB being the vertical di- ameter. Particles starting simultaneously from A, P, fall down AP, PB, respectively. Prove that the least distance between them is equal to the distance PB. MISCELLANEOUS PKOBLEMS. 165 (38) Two circles lie in the same plane, the lotvest point of one being in contact with the highest point of the,other. Prove that the time of descent from any point of the former to a point of the latter along a straight line joining these points and passing through the point of contact, is constant. (39) A particle slides from rest down a smooth sloping roof and then falls to the ground. Find the point where it reaches the gi'ound. Ans. Let I = the lengtli of the slope, a its inclination with the horizontal, h = the height of the lowest point of the slope Irom the ground. Then the distance of the point where the particle reaches the ground from the foot of the wall is 2cosa i/l sin a( ^l sin* a-{-h — ^l sin* a). (40) Two equal inclined planes are placed JxicTc to hack, and a particle projected up one flies over the top and comes to the ground just at the foot of the other. Find the velocity of projection, a being the inclination of each plane and h their common altitude. Ans. Trj/^rA (8 + cosec** ay • (41) A particle is projected from a point A with the velocity ac- quired by falling down a height a, up an inclined plane of which the base and height are each equal to b, and after quitting the plane strikes the horizontal plane AB at the point B. Find AB. Ans. AB is equal to a -|- {a'' — 6^* . (43) A particle slides doum a smooth inclined plane. Determine the point at which the plane is cut by the directrix of the path described by the particle after leaving the plane. Ans. The directrix intersects the plane at the point where the particle began its motion. (43) A particle is projected up a rough plane, inclined to the horizon at an angle of 60', ivith the velocity ivhich it would have acquired in falling freely through a space of 12 ft., and just reaches the top of the plane. Find the altitude of the plane, its roughness being such that if it were inclined to tlie horizontal at an angle of 30°, tlie particle would be on the point of sliding. Ans. 9 ft. (44) A ring slides dmvn a straight rod whilst the rod is carried uniformly in one plane, at a given angle to the horizon. Find the path described by the ring. Ans. A parabola. (45) A given circle and a given point are in the same vertical plane, the point being within the circle. Find the straight line of quickest descent from the point to the circle. Ans. The required straight line is the distance between the given point and the lower end of that chord of the circle which passes through the given point and terminates at the highest point of the circle. f46) A given point and a given straight line are in the same vertical plane. Determine the straight line of quickest descent from the given line to the point. Ans. From the given point P draw PX horizontally to meet the given line at X. Draw upwards along the line a length XFequal to PX. The straight line joining P and Y is the required straight line. 166 KINEMATICS OF A POINT — TRANSLATION. (47) To determine the straight line of slowest descent from a given point to a given circle, the point being without the circle and both in the same vertical plane, the highest point of the circle being lower than the given point. Ans. Let Pbe the given point and Q, tlie highest point of the circle. Join PQ and produce it to cut the circle at B. Then PM is the line required. (48) Determine the straight line of sloivest descent from, a given circle to a given point without it, the point and circle being in tlie same vertical plane, and the point being lower than tJie lowest point of the circle. Ans. From the given point draw an indefinite straight line cutting the circle at its lowest point, and the second intersection of the indefinite line and the circle is the required line. (49) Find the straight line of slowest descent from one given circle to another, both circles being in the same vertical plane, and each exterior to the other, the highest point of the latter circle being lower than the lowest of the former. Ans. Produce the line which joins the lowest point of the first circle and •the highest point of the second, to meet both circles again. The distance be- tween the second intersections is the required line. (50) A smooth tube of uniform bore is bent into the form of a circular arc greater than a semicircle, and placed in a vertical plane tvith its open ends upwards and in the same horizontal line. Find the velocity with which a ball that fits the tube must be pro- jected along the interior from the lowest point, in order that it may pass out at one end and re-enter at the other. Ans. If r is the radius, h the depth of the centre of the circle below the horizontal through the two ends, Vi the required velocity, «,s = |(r2 + mr + 27t2). (51) A particle slides from rest down a smooth tube in the form of the thread of a screw the axis of which is vertical. Find the time in which it voill make a complete revolution about the axis. Ans. If r is the radius of the cylinder on which the helix is described, and 4X the angle which the thread makes with the generating line of the cylinder, the required time is / 8;rr U \g sin 'Za j (52) A particle falls to the lowest point of a cycloid down any arc of the curve, the axis of the cycloid being vertical and its vertex dowmvards. Prove that the vertical velocity is greatest when it has completed half its vertical descent. (53) Also prove that it describes half the path in two thirds of the whole time. (54) If a clock pendulum lose 5 sec. a day, determine the altera- tion tvhich should be made in its length. Ans. It should be diminished by nearly the 5^-77. part of its length. (55) A seconds pendulum was too long on a given day by a quan- tity a. It was then over-corrected so as to be too short by a during MISCELLANEOUS PROBLEMS. 167 the next day. Prove that, I being the length of the seconds pendulum, the number of minutes gained in the two days was nearly 1080^. (56) A seconds pendulum earned to the top of a mountain is found to lose there 43.2 sec. a day. Find the height of the mountain, supposing the radius of the earth to be 4000 miles. Ans. 2 miles. (57) Find the time of vibration of a pendulum 20 feet long. Ans. Approximately 2.5 seconds. (58) A body dropped from the top of a wall falls to the ground while a pendulum 6 inches long makes 5 beats. Find the height of the wall. Ans. -j-Tc'^ feet. 4 (59) A seconds pendulum is lengthened one hundredth of an inch. Find how many seconds it will lose daily. Ans. About 11 seconds. (60) J^ the length of a seconds pendulum is 39.1386 inches, what will be the length of one which vibrates 40 times a minute. Ans. 88.06185 inches. (61) If the length of a seconds pendulum is 39.1393 inches, find the valu£ of g. Ans. g — 32.190 feet-per-sec, per sec. (62) A pendulum which beats seconds at the equator gains 5 mimdes a day at the pole. Compare polar and equatorial gravity. Ans. 144 to 145, approximately. (63) Two pendulums the lengths of ivhich are U and h vibrate at different points on the earth's surface. The number of vibrations which they make in the same time are in the ratio mi to m,. Find the ratio of g at the tivo places. Ans. J. (64) A seconds pendulum is carried to the top of a mountain of tchich the height w 1 mile. Find the number of seconds it unit lose daily, gravity being supposed to vary inversely as the square of the distance from the centre of the earth, and the radius of the earth to be 4000 miles. Ans. About 21.6 seconds. (65) A body revolves uniformly in a circle of 4 ft. radius in 10 seconds. Find the angular velocity and the velocity at the circum- ference. Ans. 0.628 radian per sec; 2.5 ft. per sec. (66) A body revolves uniformly in a circle of 4 ft. radium with a velocity of 8 ft. per sec. Find the normal acceleration. Ans. 16 ft.-per-sec. per sec. (67) A body revolves imiformly in a circle of 18 ft. radius. The normal acceleration is 5 ft.-per-sec. per sec. Find the velocity. Ans. 9.5 ft. per sec. 168 KINEMATICS OF A POINT — TRANSLATION. (68) A body whose velocity is 10 ft. per sec. is made to move uni- formly in a circle by a normal acceleration of 2 ft.-per-sec. per sec. Find the radius. Ans. 50 ft. (69) Two points have velocities Vi and «» and are made to move in a circle by reason of central accelerations inversely jproportional to the square of the distance from the centre. The distance of one point is ri. Find the distance ra of the other. Ans. — : — = ra* : 1 Tt ra (70) Find the relation between the distances r\ and ra and the times of revolution ti and ta. = r,« : r,«; «,* : <,« = n^ : r«» n Ta 1 I KINEMATICS OF A EIGID SYSTEM. CHAPTEK I. RIGID SYSTEM WITH ONE POINT FIXED. ROTATION. ANGULAR DISPLACEMENT. LINEAR DISPLACEMENT IN TERMS OP ANGULAR. LINE REPRESENTATIVE OP ANGULAR DISPLACEMENT. RESO- LUTION AND COMPOSITION OP ANGULAR DISPLACEMENTS. ANGULAR VELOCITY. INSTANTANEOUS AXIS OP ROTATION. ANGULAR ACCELERA- TION. RESOLUTION AND COMPOSITION OP ANGULAR VELOCITIES AND ACCELERATIONS. EQUATIONS OP MOTION OP A POINT OP A ROTATING SYSTEM. MOMENT OP ANGULAR VELOCITY AND ACCELERATION. GENERAL ANALYTICAL DETERMINATION OP RESULTANT FOR CONCURRING ANGULAR VELOCITIES AND ACCELERATIONS. Rotation. — When a rigid system moves so that all its points describe circles in parallel planes about a common straight line or axis passing through the centres of the circles and perpendicular to their planes, the system is said to rotate or have a motion of rotation about that axis. Any plane parallel to the planes of the circles is the plane of rotation. Since the system is rigid, every point must describe its circle in the same time, or the angular speed (page 72) of every point is the same. If the angular speed does not change and the plane of rotation does not change, the rotation of the system is uuiform. If either the angular speed changes or the plane of rotation changes, the rotation is variable. Motion of a Rigid System with One Point Fixed. — We have de- fined translation (page 13) as motion of a system such that every straight line joining every two points remains always parallel to itself during the motion. In such case the motion of the system is the same as that of any one of its points, and the study of the trans- * The advanced student sliould read this portion of the worl< in connection with the analogous portions of Statics referred to in the text. The student new to the subject would do well to omit this portion of the work and take it in connection with Statics later. 169 170 RIGID SYSTEM WITH OXK I'OINT FIXED— ROTATION. [CHAP. I. lation of a system is the same as the study of the translation of a point. In the preceding Chapters we have treated of the kinematics of a point or translation. If one point of a system is fixed, the motion of which it is capa- ble will be more or less complex according as its points can or can not move relatively to each other. We restrict our discussion to rigid systems, that is, systems whose points can not move relatively to each other. If one point of such a system is fixed, there can be no translation and the only motion of which it is capable is one of rotation as just defined. In such motion it is evident that all straight lines in the system must remain straight lines of unchanged length and mutual inclina- tion, and all planes must remain planes of unchanged form, area and mutual inclination. Also the motions of any two points indefi- nitely near must be indefinitely nearly the same. Angular Displacement of a Rigid System. — Let AB be the axis of rotation of a rigid system. Then, since every point must com- plete its circle in the same time, the angle de- scribed by all points in any given time must be the same as the angle described by any one point P. The angle Q between the initial and final positions, in any given time, of the perpendic- ular PO from any point Pto the axis, is called the angular displacement of the system. Since the angle 6 is measured in radians, it is independent of the length PO (page 5). Linear Displacement in Terms of Angular.— Let OPi = OPi = r be the radius for any point P which moves in a circle perpendicular to the axis at O, through the angular displace- ment PiOPa = <9, from the initial position Pi to the final position Pa. Then the triangle PiOPi is isosceles, and if we draw ON per- pendicular to P1P2 we have the linear dis- placement PiP3=2r sin 4. O Line Representative of Angular Displacement of a Rigid System. — An angular displacement of a rigid system is given when we know not only its magnitude and the direction of rotation in the plane of rotation, but also the direction of that plane in space. It is therefore a vector quantity having not only magnitude and sign, but also direction, and it can be completely represented by a straight line, like linear displace- ^ ment (page 34). Thus the length of the straight line AB denotes the magnitude of the angular displacement P1OP1 = 0. The plane of rotation is at right angles to the line AB, ivhich is therefore coincident tvith the axis of rotation. The direction of rotation is always clockwise when we look along this line in the direction indicated by the arrow. By direction of an angular displacement we mean always the direction of its line representative as denoted by the arrow. CHAP. I.] ANGULAR DISPLACEMENTS. 171 Composition and Resolution of Successive Angular Displacements. — Let a rigid system with one point fixed undergo successive angular displacements. It is required to determine the resultant angular displacement. Evidently the successive angular displacements may be about the same or about different axes, and in either case may be finite or indefinitely small. (a) About the Same Axis — Finite or Indefinitely Small. — If the axis of all the angular displacements is the same, the plane of rotation does not change, and the magnitude and sign or direction of the resultant displacement in that plane is given by the algebraic sum of the magnitudes of the successive displacements, whether they are finite or indefinitely small Inversely, any angular displacement about a given axis may be resolved into any number of successive displacements about the same axis, whether finite or indefinitely small, provided the alge- braic sum of their magnitudes is equal to the magnitude of the given displacement and the same in sign. (b) About Different Axes— Displacements Finite. — The axes must pass through the fixed point of the system. First let the displace- ments be finite. Let O be the fixed point of the system, and ORi , ORi the initial positions of the two given axes. Take ORi — ORi, and let us sup- pose first a displacement d, of the system about ORi and then a dis- placement fj2 about the new position of the other axis. During this motion R2 and Ri will move on the surface of a sphere. When the system is rotated an angle 9i about ORi, the axis ORt moves from ORi to OR^'. Now join R1R2, R1R2' and RiR^ by great circles of the sphere. Then the angle R^RiRi' = Oi. Bisect this angle by a great circle meeting Ri'R-. at D. Draw a great circle through R^' inclined to Ri'Ri at the angle ^ and meeting RiD at R. Then draw Ri'Ci, making the same angle ^ with R^'Rx on the other side, and make R^Ci = R^R. Then i2,a will equal RxR , and the angle Ri'RiC-i = ^■ When then the system is rotated about ORi , and the axis OR^ moves to OR/ through the angle R^RM/ = fii, the Ime OR will move to OC2 through the angle RRxd = 0i. If now the system is rotated about OR2 through the angle CiR'R = 0-2, the line OC2 moves back to OR. Hence tho line OR has the same position before and after the rotations. The resultant displacement is then a displacement about Hence, the resultant of two successive rotations 6, about ORi and «2 about OR2 , when the axes intersect in anoint O, is a single rota- tion e about the axis OR passing through 0. In order to find the position of this axis OR and the ma^itude of 0, we have in the spherical triangle RiRRi the angle RKxKi - -Oi, the angle RR^R = ^6, and the exterior angle DRR^ = :^^- Hence 173 EIGID SYSTEM WITH ONE POINT FIXED — ROTATION. [CHAP. I^ COS -6 = COS -9, COS -02 — sin -6, sin -62 cos RiBi ; . (1> sin RiR sin RRi sin RiRi sin -Q2 sin -61 sin -6 2 2 2 (2) Since OJRi and OR2 are lines of a rigid body and ORi coincides with the position of the first axis of rotation in space, the second axis of rotation in space has the position OR-i' and not OR2* Hence, in general, the order of the two successive rotations is not indifferent. Example. — The telescope of a theodolite, originally horizontal and pointing north, is first turned into an altitude of 60° and then turned towards the west into the prime vertical. Find the resultant rotation. Ans. We have Qj = 60°, 62 = 90°, BiB^ = 90°. Hence cos 1 = y 8 X |/| = i/|, or Sin i» = i/|. For the position of the axis we have itB= ^=24/-^, sini2i?2 = ^ = 4/%. 2^2 2^2 sini2i If we invert the order of the two rotations, we have 9i = 90°, 62 = 60" B1B2 = 90°. Hence cos -0 = g-y ^' or sin -9 = 2^2' ^ ^^^^^^ ' For the position of the axis sin BiB = — — = i/i, sin BB^ = — = 2 a/I. 2 r ^ 2 r 2 (c) About Different Axes — The Displacements Indefinitely Small. — A ^— ^^c ^®cond let the rotations be indefi- K/^ ^„.,-''^/ nitely small. Let OA = 9 and OB y^ ^ ^X = be the line representatives. Com - /y^-^"^^"^ ''' plete the parallelogram and draw -,^d_5_ ^''' ^G- Let P be any point of the sys- " s B tern in the plane of OA and 05, and draw the perpendiculars P§, PR, PS. When rotation occurs- about OA, the point P will move perpendicularly to the plane of the paper through a very small distance represented by rO or OA x PQ' (page 5). When rotation occurs about OB, the point P moves perpendic- ularly to the plane of the paper also, through a very small distance represented by OB x PS. Since both these displacements are very small they coincide in direction, and the resultant is OAx PQ + OB X PS=OC X PR. Hence the resultant displacement is given by OC. •€HAP. I.] AXIS OF ROTATION. 173 We should have the same result if the rotation about OB occurred first, also if the point P had been taken within the angle AOB. Also whether OA and OB are axes fixed in the body or in space. If we have more than two successive rotations, the third may be compounded with the resultant of the first two in like manner. Hence if a rigid system with one point fixed undergo any num- ber of successive indefinitely small angular displacements about different axes either fixed in the system or fixed in space, the resultant angular displacement is obtained by treating the line representatives precisely like linear displacements (page 35). We have thus the parallelogram and polygon of angular displace- ments. Composition and Resolution of Simultaneous Angular Displace- ments. — The simultaneous angular displacements may be finite or indefinitely small and must be either about the same axis or differ- ent axes. (a) About the Same Axis.— If the axis of all the angular displace- ments is the same, the plane of rotation does not change and the magnitude and sign or direction of the resultant displacement in that plane is given by the algebraic sum of the magnitudes of the simultaneous angular displacements, whether they are finite or in- definitely small. (b) About Diflferent Axes. — If the rotations OA, OB are indefi- nitely small, we see from the figure, page 171, that it makes no differ- ence whether they are successive or simultaneous. We can resolve and combine them, therefore, by their line representatives just like linear displacements (page 35). We have then the parallelogram and polygon of angular displacements. If the rotations OA, OB are finite, we can divide each up into a number of indefinitely small rotations and treat each pair as before. We have then the parallelogram and polygon of angular displace- ments in this case also. Composition and Resolution of Angular Displacements in gen- eral. — We see then that in all cases except finite successive angular displacements about different axes we can combine and resolve any number of angular displacements whether simultaneous or successive, finite or indefinitely small, about the same or about different axes by means of the line representatives, just like linear displacements. Simultaneous angular displacements are usually called com- ponent angular displacements. Component angular displacements must then be understood to always mean simultaneous angular dis- placements, unless otherwise specified. Sign of Components of Angular Displacement. — The sign of the line representatives of the components along the axes X, Y, Z of an angular displacement follows the same rule as for linear displace- ment fpage 36). Hence if we look along the line representatives towards the origin, the radius vector wUl always be seen to move connter-clochwise. Axis of Rotation.— In every possible dis- placement of a rigid system with one point fixed, there is one line fixed in the system pass^ ing through the fixed point, called the axis <^ rotation, which has the same position in both the initial and final positions of the system. Let O be. the fixed point of the system, and . . ._ let A,, Bi be the initial and Aa, B^ the final positions m space of two points of the system. 174 RIGID SYSTEM WITH ONE POINT FIXED — ROTATION. [CHAP. I^ Since the system is rigid, OAi = OA^ , and OBi — OB 2. Let Ai be brought to Ai by rotation about an axis through O perpendicu- lar to the plane of AiOAt. By this rotation Bi moves to 6, and since the system is rigid, Oh = OBi —OB^, and A^b — A2B2. The triangles OAib and OA-iB^ are then equal in all respects, and 6 can be brought to JSa by rotation about OA-i. The given displacement can always then be produced by two successive rotations about two axes passing through O. As we have seen (page 171), two such successive rotations give as resultant a single rotation about an axis through O. This is the axis of rotation. Cor. 1. Hence any angular displacement of a rigid system with one point fixed is completely specified by the line representative of the resultant angular displacement, which coincides in direction with the axis of rotation. CoR. 3. Any angular displacement of a rigid system with one point fixed may be resolved into three angular displacements about the co-ordinate axes through the fixed point taken as origin. Cor, 3. Every line in the system parallel to the axis of rotation remains unchanged in direction. Mean Angular Velocity of a Rigid System. — The magnitude of the angular displacement during a given time of a rigid system with one point fixed, divided by the number of units of time, gives the magnitude of the mean angular velocity of the system. It is represented by a line just like angular displacement (page 170). By direction of mean angular velocity we always mean direc- tion of the line representative. Mean angular speed then is mean time-rate of angle described (page 72). Mean angular velocity is mean time-rate of angular displacement. Instantaneous Angular Velocity of a Rigid System. — The limit- ing magnitude and direction of the mean angular velocity when the interval of time is indefinitely small is the instantaneous angu- lar velocity. The term angular velocity always signifies instantaneous angu- lar velocity unless otherwise specified. It may be represented by a straight line just like angular dis- placement (page 170;. By direction of an angular velocity we always mean the direc- tion of its line representative. We see then that angular displacement and angular velocity are vector quantities like linear displacement and linear velocity. Angular velocity is directed angular speed, just as linear velocity is directed linear speed. Speed is magnitude of velocity, whether linear or angular (page 43). Instantaneous Axis of Rotation. — The instantaneous angular velocity of a rigid system is then given by its line representative. This line representative coincides in position with the axis of rota- tion at the instant. This axis is then the instantaneous axis of rota- tion. Unit of Angular Velocity. — Since the magnitude of the angular velocity at any instant is the angular speed in a given direction at that instant, the unit of angular velocity is the same as for angul; r speed, or one radian per sec. We denote the magnitude then by the same letter, oa, and we have the same numeric equations as for angular speed (page 73). Thus for mean angular velocity CO = T 1 W CHAP. I.] ANGULAR ACCELERATION. 175 and for instantaneous angular velocity - = * («) TTnifonn and Variable Angular Velocity. — Angular velocity is uniform when the line representative has the same magnitude and direction whatever the interval of time. Uniform angular velocity is then uniform angular speed in an unchanging plane, just as uni- form linear velocity is uniform linear speed in an unchanging direction (page 43). In such case angular velocity is the same as the mean angular velocity for any interval of time. When eivher the magnitude or direction of the angular velocity changes it is variable. W hen the magnitude alone changes we have variable angular speed in an unchanged plane of rotation. When the direction only changes we have uniform angular speed in a changing plane of rotation. When both change we have variable angular speed in a changing plane of rotation. Mean Angular Acceleration of a Rigid System. — If OA = co, and OB — CO are the line representatives of the initial and final angular velocities of a rigid system with one point fixed, during any time t, then AB is the line represent- ative of the integral angular acceleration of the AB system during the time t, and —r- gives the magnitude of the mean angular acceleration whose direction is AB. (Compare page 48.) Mean angular acceleration then is time-rateof change of angular velocity, whether that change takes place in the direction of tJie an- gular velocity or not. Instantaneous Angular Acceleration of a Rigid System. — The limiting magnitude and direction of the mean angular acceleration when the interval of time is indefinitely small is the instantaneous angular acceleration. It is the limiting time-rate of change of an- gular velocity whether that change takes place in the direction of the angular velocity or not. Angular acceleration always signifies instantaneous angular acceleration unless otherwise specified. , j. It may be represented by a straight line just hke angular dis- placement (page 170). By direction of an angular acceleration we mean the direction of its line representative. Instantaneous Axis of Angular Acceleration.— The instantaneous angular acceleration of a rigid system is then given by its line rep- resentative. This line representative coincides m position witn the axis of angular acceleration at the instant. This axis is then the instantaneous axis of angular acceleration. Angular acceleration may be zero, uniform or variable. VV hen it is zero, the angular velocity is uniform and we have uniform angular speed and an unchanging plane of rotation. When It is uniform, it has the same magnitude and the same direction whatever the interval of time. In such case the accelera- tion is equal to the mean acceleration for a^J .^"t^rval of time It the direction coincides with that of the i^^t^^l velocity, we have uniform rate of change of angular speed and ^n unchanged plane of ro:ation. If it makes an angle with the velocity, we have a changing plane of rotation and variable velocity. 176 RIGID SYSTEM WITH ONE POINT FIXED — ROTATION. [CHAP. I. When it is variable, either direction or magnitude changes or Tjoth change. If the angular acceleration is always at right angles to the an- gular velocity, it only changes the direction but not the magnitude of that velocity. Hence, just as on page 53 a normal linear acceleration has no effect upon the linear speed, but only changes the direction of mo- tion, so, if a rigid system rotating with given angular speed about an axis has an angular acceleration about an axis always perpendic- ular to the first, there is change of direction of this axis but no change of angular speed about it. The gyroscope is an illustration of this principle. Resolution and Composition of Angular Velocity and Accelera- tion. — Since for an indefinitely small time the angular displace- ment is indefinitely small, we see from page 171 that we can com- bine angular velocities and accelerations, whether simultaneous or successive, by means of their line representatives just like linear velocities and accelerations (page 43 j. Sign of Components of Angular Velocity and Acceleration. — The sign of the line representatives of the components along the axes X, Y, Z of an angular velocity or acceleration follows the same rule as for linear velocities and accelerations (pages 44, 50). Unit of Angular Acceleration. — Angular acceleration is meas- ured in terms of the same unit as rate of change of angular speed (page 73), or one radian-per-sec. per sec. We denote its magni- tude then by the same letter, ex. Relations between Angular and Linear Velocity and Accelera- tion. — We have also the same relations between angular and linear acceleration and velocity as for a point moving in a circle (page 76). Thus we have, for any point of a rigid system whose distance from the axis of rotation at any distance is r, roo^=v, ra=ft, voo =fn = roo^ = —, fp=ftr = r^a, vr = r'ao. Equations of Motion of a Rotating Rigid System under Diflferent Angular Accelerations. — Since angular velocities and accelerations are represented by straight lines, just like linear velocities and ac- celerations, we have the same equations for motion of a rotating rigid system as on page 50. We have only to substitute go for v, Q for s, a for /. With these substitutions equations (1) to (14), page 50, hold good and it is unnecessary to repeat them here. Moment of Angular Displacement. — Just as we called the prod- uct of the magnitude of a linear displacement by the magnitude of the perpendicular let fall from any given point upon its du-ec- tion the moment of the linear displacement (page 60), so for angular displacement we call the product of its magnitude by the magnitude of the perpendicular from any point upon the direction of the line representative the moment of the angular displacement We take its sign just as for moment of linear displacement, page 62. Since the line representative is coincident with the axis, the perpendicular is the distance of the point from the axis. Thus if AB = 6 is the line representative of an angular dis- placement OiOOi = of a rigid system, the axis has the position CHAP. I.] CONCURRIN-G ANGULAR DISPLACEMENTS, ETC. 177 AOB. If then Oi is the initial position of any point of the sys- tem and OOi = p is the perpendic- ^ ular from Oi upon the axis or direc- tion of the line representative AB, the moment is ± p'j according to direction, just as for moment of linear displace- ment (page 62;. But pO is the length j^ , ^ of the arc OiOi described by the point ' -* ' ^^^ Oi in a plane perpendicular to the plane of AB and 00 1. Hence, the moment pS of the angular displacement of a rigid system rela- tive to any point of the system gives the length of the arc O1O2 described by that point in a plane perpendicular to the plane of the axis AB and the radius vector p. The corresponding linear displacement of Oi is evidently T = 2p sin- (1) Since the angle OO1O2 equals the angle OOaOi, we have for the direction of the linear displacement relative to OOi , angleOOiO. = 90°- 0.0r=^^ (2) We have also, just as on page 62, the algebraic sum of the moments of any number of component angular displacements, rel- ative to any point, equal to the moment of the resultant. Also, just as on page 60, the line representative of an angular displacement may be laid off from any point in its line of direction without affecting its moment. Moment of Angular Velocity or Acceleration. — Just as we called the product of the magnitude of a linear velocity or accelera- tion by the magnitude of the perpendicular from any given point upon its direction the moment of the linear velocity or acceleration, so for angular velocity or acceleration we call the product of the magnitude by the magnitude of the perpendicular from any point upon the direction of the line representative the moment of the angular velocity or acceleration. We take its sign just as for moment of linear velocity or accele- ration (page 60). Since the line representative is coincident with the axis, the perpendicular is the distance of the point from the axis. Thus if AB = (a is the line representative of an angular velocity of a rigid system, the instantaneous axis has the position AOB. If then Oi is any point of the system and OOi =p is the perpendicu- lar from Oi on the axis or direction of the line representative, the moment is ± poo according to direction, just as for moment of linear velocity (page 60). But this is ^^■^ the linear velocity v of Oi at the instant, in a direction perpendicular to the plane of AB and 00 1. Hence, the moment poo of the angidar velocity co of a rigid system, relative to any point of the system gives the, linear velocity V of that point in a direction perpendicu- lar to the plane of the instantaneous axis of rotation AB and the instantaneous radius vector p. 178 BIGID SYSTEM WITH ONE POINT FIXED — KOTATION". [CHAP. I^ In the same way, the moment pa of the angular acceleration a of a rigid system relative to any point of the system gives the linear- tangential acceleration ft of that point in a direction perpendicular- to the plane of the instantaneous axis of angular acceleration and the instantaneous radius vector. We have also, just as on page 62, the algebraic sum of the mo- ments of any number of component angular velocities or accelera- tions, relative to any point, equal to the moment of the resultant. Also, just as on page 60, the line representative of an angular velocity or acceleration may be laid off from any point in its line of direction without affecting its moment. Concurring Angular Displacements, Velocities or Accelerations. — "We see then that angular displacements, velocities or accelera- tions are represented by straight lines, called line representatives^ which coincide with the axis of rotation. We deal with them entirely by means of these line representatives. When we speak of their "direction," we mean the direction of the line representa- tives. We resolve and combine them by means of their line repre- sentatives, and in the same way we have their moments just as for linear displacements, velocities or accelerations. Following the same analogy, we can speak of them as "applied " or "acting" at. certain points. When they all intersect at the same point, we may call them concurring, just as if they were linear. When they do not intersect at the same point they are non-concurring. When they act in the same direction in the same line they are conspiring. When in the same or opposite directions in parallel lines they are parallel. When in opposite directions in the same line or in paral- lel lines they are opposite. When they lie in the same plane they are co-planar. Condition for Rotation only. —If a rigid system has one point fixed, it can have no translation but only rotation, and therefore all the component angular displacements, velocities or accelerations must reduce to a concurring system, so that we have a single result- ant angular displacement, velocity or acceleration about an axis through this point, which is therefore at rest. General Analytical Determination of Resultant Angular Dis- placement, Velocity or Acceleration for any Number of Concurring Components. — We see then that all the equations of pages 63 to 65 hold good for angular displacements, velocities or accelerations, as well as for linear. For angular displacements we have only to substitute 6 in place of V. The moments Mx, My, Mz then give the arcs of displacement about the axes of X, F, Z of the origin, considered as a point of the rigid system, rotating about the resultant axis. For angular velocities we have only to substitute oo for v. The moments Mx , My, Mz then give the component linear velocities Vx , Vy , Vz along the axes of X, F, Z of the origin, considered as a point of the rigid system, rotating about the resultant axis. For angular accelerations we have only to substitute a for u. The moments Mx, My, Mz then give the component linear tangential accelerations ftx, fty, ftz along the axes of X, Y, Z of the origin, considered as a point of the rigid system, rotating about the re- sultant axis. To make our notation consistent we should also replace cos a,, cos h, cos c, page 65, by cos d, cos e, cos /, and replace cos d, cos e» cos f, page 66, by cos a, cos 6, cos c. We have then from page 65, equation (4), for the component. linear velocities Vx,Vy, Vz along the axes of X, Y, Z of the origiUy CHAP. I.] TWO CONCUERING ANGULAB VELOCITIES. 179 considered as a point of the rigid system, rotating about the re- sultant axis, Vx = oozy — aoyZ ; Vy = OOxZ — 00zX\ Vz = Oi>yX — COxy- We have also in the same way fix = azy — ayZ ; fty = axZ — azX ; ftz = cXyX — axy. (1) (2) Equations (2) give the component linear tangential accelerations along the axes of X, Y, Z of the origin, considered as a point of the rigid system, rotating about the resultant axis. If we multiply the first of equations (1) by eox , the second by coy , the third by a>z and add, we obtain VxOOx + VyOOy + VzOOz = 0. (3) Equation (3) is the condition for rotation only. When it is ful- filled, we know that the motion of the system is that of rotation only about the instantaneous axis. Resultant of Two Concurring Component Angular Displacements, Velocities or Accelerations.*— It will be of profit to specially discuss the case of two concurring component angular displacements, velocities or accelerations. Let the two angular velocities oji, GOi be in the same plane and pass through the points A and B of a rigid system, so that they intersect at O. Then the resultant oo,- must pass through O and be in the plane of CiJi, ooi. Take any point P in this plane and draw the perpendiculars Pwi = p,, Pn.=pi, Pn=pr. Then, since the moment of the resultant is equal to the algebraic sum of the moments of the components, GOrPr = 0Oipi + OOiPi, . (1) where regard must be paid to the signs in any case. Thus we have in the figure 00,p,- = QOipi — OO^Pt. Draw the line AB, intersecting the resultant oor at the point C Let <:i-i be the angle of «,, and a^ the angle of <a, with AB. If we take moments about C, we have oo^ . ACsinai = oo^ .EC sin a-,. * Compare Statics— Non-concurring Forces. 180 KIGID SYSTEM WITH ONE POINT FIXED — ROTATION. [CHAP. I. But AC + BC=AB. Hence AC— <"3 • ^-Bsin tj „^ _ 00, .AB sin a, ,„, (Uisinai + ftjasinaj' ajisinai + (Wasinaa We thus know the position of the resultant go,- in the plane of ooi and ooj. Magnitude and Direction of the Resultant.* — If we lay off in Fig. 3, ftJi and ooi, then, just as for linear velocities, OC = oor gives the magnitude and direction of the resultant. Take rectangular axes OX, OY, Fig. 1, in the plane of go,, go2, and let OX be parallel to AB. Let go, make the angle a, with OX and /3i with OY, and oo^ make the angle a^ with OX and fJ-i with OY. Denote the algebraic sum of the components parallel to OX by oox and parallel to OF by ooy. Then we have cox = GOi cos a, + GOi COS aa; | GOy = COi COS pi + CJa COS Pa; J where we must pay regard to signs. Thus components in the direction OX, O Y are positive, in the opposite directions negative. If the resultant oor makes the angles d and e with the axes of X and Y, we have cos d — — cos e = -^ (4) Squaring and adding, GOr = ^ QOx + Oiiy (5) The magnitude and direction of the resultant are thus deter- mined. Also if ^1 is the angle of go, with the resultant, and % the angle of GiJa with the resultant, and & the angle between w, and oot , we have directly from Fig. 2 sin 01 = — sin 0, sin Oj = — sin 0, (6) and OOr = \/cOi^ + CJa* ± 200\00i COS 9, (7) where the ( + ) sign is used when Q is less than 90°, and the (— ) sign when is greater than 90°. The tangent of the angle d which the resultant makes with AB or OX is tand = -'' (8) GOx From (6) and (7) we can find the magnitude and direction of the resultant directly if 6i is known. If a, and aa are known, (3) and (5; give wr , and (4) or (8) the direction. From (1) we have also GOr where regard must be had for the signs of oo,p, and oo^p-i in any case. From (9) for any given point P for which p, and p^ are known, we can locate the resultant by describing a circle with centre P and radius pr and drawing GOr tangent to this circle in the direction given by (6). * Compare Statics. CHAP. I.] TWO PARALLEL AXGULAR DISPLACEMENTS, ETC. 181 The same formulas hold good for two concurring angular ac- celerations. We have only to replace go by a. [The student will of course not confuse this a, which stands for angular acceleration, with ci-i , cXi in the formulas, which stand for angles.] The same formulas hold good also for two concurring component angular displacements. We have only to replace cj by d. When the Angular Displacements, Velocities or Accelerations are Parallel. — In this case ai and aa are equal, S = 0, the intersection O is at an infinite distance, o^r = <»i + oja , and we have from (,2) AC=~ . AB, BC=~.AB; 0) and hence, multiplying the first by ooi and the second by goq , Goi . AC = &3q . BC, or — = -—, 002 AC To prove this independently, take C as the point of moments (2) Then whether the line representatives act in the same or in opposite directions, we have GO,pi — GOiPi = 0, or GOipi = COiPi. But from similar triangles », AC ^ ooi BC — = ^-, hence — = ^^. Pi BC coi AC The same holds for angular displacements or accelerations. We see from (1) that the distances AC and BC depend only upon the magnitudes of goi and co^ and the distance AB, and not at all upon the common direction of ooi and 002. Therefore if ooi and <»j always pass through the points A and B no matter what their common direction, the resultant aor always passes through C. The point C is then the point of application of the resultant cor for aU directions. . 1 j- Hence, the resultant of tivo parallel component angular dis- placements, velocities or accelerations is in their plane and equal in magnitude to their algebraic sum. It acts parallel to the com- ponents in the direction of the greater. If the components always pass through tivo given points A and B, the resultant always passes through a point C no matter tvhat the common direction. This point C is then the point of application of the resultant. It is on the straight line AB or this line produced, and dimdes it into seg- ments inversely as the components. Or tlie products of the com- fonents into their adjacent segments are equal. (Compare btatics— 'arellel Forces.) 182 RIGID SYSTEM WITH ONE POINT FIXED — ROTATION. [CHAP. I, Cor. 1. When the components act in the same direction, the resultant lies within the components and nearest the larger. When the components act in opposite directions, the resultant lies without the components and on the side of the larger. CoR. 2. When the components are opposite and equal in magni- tude, oor = 0. Also from (1), AC =ao, BC = ao, or the resultant is zero and acts at an infinite distance. That is, equal and opposite parallel components cannot have a single resultant. Such a system is called a coapl6. (Compare Statics— Parallel ^Forces.) EXAMPLES. tl) A rigid system has two component rotations of 2 and 4 radians about axes inclined 60°. Find the resultant rotation. Ans. Component rotations are understood to be simultaneous unless other- wise specified (page 172). Hence magnitude of resultant rotation is 2 4/7 radians; axis inclined at an angle with the greater component whose sine is 4/3 2|/7' (2) A sphere with one of its superficial points fixed has two com- ponent rotations — one of 8 radians about a tangent line and one of 15 radians about a diameter. Find the axis of the resultant dis- placement and the number of complete revolutions made about it. Ans. Inclination of axis to greater component at an angle whose tang is —r. Resultant displacement 17 radians, number of complete revolutions ;r-. 15 27t (3) A sphere is rotating uniformly about a diameter at the rate of 10 radians per min. Find (a) the component angular velocity about another diameter inclined 30° to the former, and (b) the com- ponent rotation produx;ed in 2 min. about a diameter inclined 45° to the first. Ans. (a) 5 4/3 radians per min. ; (6) 10 4/2 radians, (4) A pendulum suspended at a point in the polar axis of the earth oscillates in a vertical plane. Find the motion of this plane relative to the earth. Ans. The plane of the pendulum is fixed in space, and the motion of the earth with reference to this plane is a rotation from west to east at the rate of one revolution per day. The motion of the plane relative to the earth is then from east to west at the same rate of one revolution per day. (5) A pendulum is hung at a place of latitude A and oscillates in a vertical plane. Find (a) the angular velocity of the plane of the pendidums motion relative to the earth, and (6) the time in xvhich this plane ivill make one complete revolution at a place in latitude 60° iV. Ans. The angular velocity of the earth about its axis is 27f radians per day. The component of this in the direction of an axis through the centre of the earth and the point of suspension of the pendulum is 2n sin A radians per day from west to east. This is the motion of the earth relative to the plane of the pendulum. Hence — (tl) The motion of the plane of the pendulum relative to the earth is 2it sin \ -radians per day from east to west; -CHAP. 1.] EXAMPLES. 183 (6) The time of revolution is — — — - ^ - _?_ d„v8 aff sin A sin 60 ~ ^/^ *^®' (6) A cube rotates about a vertically upward axis through one of tts edges. At a given instant at which the diagonal of the upper surf ace passing through the axis points north the cube has an an- gular velocity of iO radians per sec, and begins to have a uniform angular acceleration about an axis vertically dowmvards through the same edge of 6 rad.-per-sec. per sec. Find (a) the direction in which the diagonal will point after 20 sec. ; (6) the number of revolu- tions made by the cube. Aus. From the equations of motion page 73 we have co = coi — at, 6 — Qi = ooit — 2^^^- ^® ^^v® '"i = 40 radians per sec., a = 6 rad.-per-sec. per sec, t = 20 sec, 8i = 0. GO ^0 In the time ti = — = ^ sec. the cube comes to rest and has the angular displacement 9i = -^ radians towards the east. o It then moves in the opposite direction towards the west durinff the time . «n 20 40 , , , ^1 t = aO — Q ~ IT ^®^- ^^" undergoes the angular displacement = —at^ = —^ radians. Hence (a) the angular displacement from the north point to- Avards the west is 400 radians or 63.661 revolutions, or 63 revolutions and 26TM W. The direction of the diagonal is then S. 57°.96 E. (b) The total 2000 angular displacement is — ^ — radians, hence the number of revolutions is 333.3 (7) A sphere is rotating at a given instant about a given diameter ACB with an angular velocity of 4 rad. per min. It has an angular acceleration of 2 rad.-per-min. per min. about a diameter inclined 30° to ACB. Find (a) the angular velocity, and (6) the angular dis- placement after 20 min. Ans. (Page 174.) (a) 4v 101 + 10 ^S rad. per min. inclined to CB at an angle whose tangent is -= ; (b) 80 y 26 + 5 |/3 radians inclined to CB 5 at any angle whose tangent is -=. <& -|- o y o (8) A rigid sy stein has one point fixed. The co-ordinates of this Xyoint xvith reference to any point of the system taken as origin are at any instant x= +B ft., y = + 4 ft., z = 0. The component an- gular velocities at this instant are sji = 40, odi ■= 50, oo^ = 60 radians X)er sec, the line representatives making the angles ai = 60', /^i = 150% r. = 90°; a2 = 120% ^^ = 30% r^ = 90'; as = 120% ^3 = 150% r^ = 90\ Find the resultant angular velocity. Ans. (See Example (1), page 67.) The component angular velocities are in one plane and fija; = — 35 radians per sec, GOy=— 43.3 radians per sec. The resultant is oo,- = 55.67 radians per sec, its line representative or the instantaneous axis of rotatioii making with the horizontal the angle d — 128° 67' 17", and with the vertical the angle e = 141° 2' 43". If we look along this 184 KINEMATICS OF A RIGID SYSTEM — ROTATION. [CHAP. I, line representative wliicli passes through the fixed point, towards the origin, the rotation will be seen as counter-clockwise. The moment of the resultant angular velocity oor with reference to the point gives us the linear velocity of rotation at about the instantaneous axis, «j, = -f 10 ft. per sec. in a direction through perpendicular to the plane XT, or along Z, from towards Z. The distance of from the axi^ is p = about 0.18 ft. The equation of the axis is y = 1.237a! -|- 0.286. Its intercepts on the axis are / = + 0.286 ft., x - - 0.232 ft. (9) Express and solve the same example for component angular accelerations and displacements. (10) A rigid system has one point fixed. The co-ordinates of this point with reference to any point of the system taken as an origin are at any instant x = + 3 ft., y=+ ^ft., z= +5ft. The compo- nent angular velocities at this instant are goi — 40, cot = 50, oo-i — 60 radians per sec., the line representatives making the angles with the axe4 o-j = 60°, (ii = 100°, yi obtuse ; a^ = 100°, /Sj = 60°, y^ acute; as = 120°, fJa = 100% ya acute. Find the resultant angular velocity. Ans. (See Example (2), page 67.) We have oox=- 18.6824, GOy=-{- 7.635, ooz = -{- 59.391 radians per sec. The resultant angular velocity is cor = 62.73 radians per sec, its line rep- resentative making with the axes the angles d = 118° 17' 33", e = 85° 6' 12", / = 13° 30' 24". This line representative passes through the fixed point and gives the in- stantaneous axis of rotation. If we look along this line towards the origin, the rotation will be seen as counter-clockwise. The velocities of rotation at along the axes are rz = + 97.6346, vx = -^ 199.389, % = - 371.585 ft. per sec. The resultant velocity of rotation at is «r = 407.6 ft. per sec, making •with the axes angles a = 60° 42' 57", b = 131° 46' 34", c = 76° 8' 31". The equations of the projections of the axis upon the co-ordinate planes are: on plane XT, y = - 0.408a! -f 5.226 ; " '• TZ, s =+7.778^^-26.115; " " ZX, x= - 0.314s -\- 4.572. The axis pierces the plane XT at «' = + 4.572 ft., / = + 3.357 ft.; " " '= " " TZ " y' = -f 5.226 ft., 2' = - 14.56 ft.; «• •• " " " ZX " s' =- 26.115 ft., «' = + 12.809 ft. (11) Express and solve the same example for component angular accelerations and displacements. (12) Let the axes of tivo concurring angular velocities of a mgid sy stern, gdi = 20, ooi = 30 radians per sec, pass through the points A, B of the system, the distance AB = 2 ft., and the angles a^ = 60°, a^ = 30°. Find the point C on the line AB through which the resultant axis passes, and the magnitude and direction of the resultant angu- lar velocity. CHAP. I.] EXAMPLES. 185 Ans. AC = 0.928 ft., ajx = + 15.98 radians per sec, GOy= — 32.33 radians per sec. The angle of the resultant with AB is given by 32 32 tan d = - j^ = - 2.022, or d = 63° 41' = BCcor. The resultant is GOr = 36.05 radians per sec. We have also for the angle of the resultant with t»i , since 6 = 90", on 8i» 6, = sl^ = 0.832, or 6, = 56° 19'. 00.05 (13) Express and solve the same example for component angular accelerations and displacements. CHAPTEE IL MOMENT OF A COUPLE. DISPLACEMENT OP A RIGID SYSTEM. RIGID PLANE SYSTEM. COMPOSITION AND RESOLUTION OF TRANSLATION AND ANGULAR DISPLACEMENT. COM- POSITION AND RESOLUTION OP TRANSLATION AND ANGULAR VELOCITY. CENTRAL AXIS. SCREW MOTION. ROTATION AND RECTILINEAR TRANS- LATION. COMBINED PARALLEL ROTATIONS, ONE AXIS FIXED. INTER- SECTING AXES, ONE AXIS FIXED. ANALYTIC DETERMINATION OF RE- SULTANT ANGULAR VELOCITY AND VELOCITY OF TRANSLATION FOR NON- CONCURRING ANGULAR VELOCITIES. Moment of a Couple.* — We have just seen in the preceding Chapter, page 181, that two parallel equal and opposite components acting at different points of a rigid system constitute a couple. We may have then an angular-displace- ment couple or angular- velocity cou- ple or angular-acceleration couple. Let + 00, — CO, acting at the points A, -B of a rigid system, constitute an ~Q, angular-velocity couple. If we take any point C between the components, or any point d , d on either side, in the plane of the components, we have in the first Ci case, denoting the distance AB by p, for the moment about C, just as for linear velocities (page 60), -a>.AC-oo.BC=- oo{AC + BC) = -poo. In the second case, for the moment about Ci we have 00 . CiA - 00. aB = - co{aB - C^A) = -poo. In the third case, for the moment about d we have — CO . CiA + 00 . CiB = — Go{CiA — CiB) = — pan. Hence the moment about every point in the plane of the couple is constant and equal to ± poo, the (-1-) or (— ) sign denoting direction just as for moment of linear velocity (page 60). For an angular-acceleration couple we have in the same way ± pcx, for an angular-displacement couple ±jp6. We see then that the moment of a couple is the same for every point in its plane and equal to the product of either of the compo- nents by the distance between them. * Compare Statics— Parallel Forces. 186 CHAP, n.] DISPLACEMENT OF A RIGID SYSTEM. 187 a.-*^' Composition and Resolution of Translation and Angular Displace- ment.— Let a rigid system have a rotation of OiOO^ = Q radians about an axis AOB through the point O, and AB = 9 be the line representative. If we take any other point of lae system, as Oi , and at this point apply the two equal and opposite angular displacements Oia — —B and Oih = + 6, both parallel to AB, it is evident that the motion of the system is not affected. We have then the angular displace- ment about the axis AOB reduced to an equal angular displace- ment 0\b about a parallel axis through Oi and a couple AB and Om. The moment of this couple is the same for every point in its plane and equal to pO, where p is the perpendicular distance be- tween the components AB and Om of the couple. But we have seen (page 177) that the moment pO corresponds to a linear displacement in a plane perpendicular to the plane of the couple of T = 2p sin -, making an angle 00x0% withjp given by Tt—B 00.0% = (1) (2) Hence, an angular displacement Q about any given axis can be resolved into an equal angular displacement aboid a parallel axis through any point of the system and a linear translation in the plane of rotation of the system whose magnitude and direction are given by (1) and (2). Conversely, the resultant of the rotation of a rigid system about a given axis and a translation in any given direction, is an equal rotation about a parallel axis, whose position with reference to the first can he determined by (1) and (2). Cor. 1. Two non-concurring angular displacements can be re- duced to a resultant angular displacement about a resultant axis at any point and a couple which causes translation. Cor. 2. Hence if we have any number of component angular rotations about any axes, whether these axes intersect or not, we can reduce each to an equal rotation about a parallel axis through some one point of the system and a translation of the system. The resultant translation can then be found as on page 35, and I he resultant rotation as on page 173, for simultaneous angular dis- placements. Cor. 3. Therefore any number of component translations and rotations can all be reduced to a single translation and a single rotation about any given point. It is evident that this single rota- tion is not affected by the position of the point, which affects the translation only. Displacement of a Rigid System. — Any displacement of a rigid £ystem may be produced by a translation and an angular displace- ment. 188 RIGID SYSTEM— TRANSLATION AND ROTATION. [CHAP. II» •C. Let Ai,Bi, Ci be the positions of any three points which determine the initial position of the rigid system. Let A-2 , Bi, Ci be the final position of these jjoints after any displacement. First let the system be translated, sa that Ai comes to its final position Aj. Then Bi and Ci will take the positions b and c, the lines Bib and CiC being equal and parallel to AiAi. We see then that Ai is a fixed point in the system so far as the two positions AiBiCi , AiBiCi are concerned. But we have seen (page 173) that in every possible displacement of a rigid system with one point fixed there is an axis of rotation fixed in the system which remains unchanged. Hence A^cb can be brought to the position A^BiCi by rotation about that axis. OoK. i. It follows that the displacement of a rigid system is known if the magnitude and direction of the linear displacement of any point is known, and also the magnitude and direction of the angular displacement of the system about that point. Cor. 2. Also, the displacement of a rigid system is known if the magnitude and direction of the component linear displacements of any point parallel to three rectangular axes and of the component angular displacements of the system about axes parallel to the first through the point are known. Rigid Plane System. — Any displacement of a rigid plane system in its own plane may be produced by rotation about some point in the plane. Let AiBi and A-^B^ be the initial and final positions in the plane, of the same line of the system, o ^ so that AiBi and A^Bi are of \\ns. -^- -,r equal length. Join AA-i and w N^^ ^--- ^^- a BiBi by lines and bisect these lines at C and D. Erect per- Eendiculars at the points of \ \ \ ^"-x^ 7d isection C and D and produce them to intersection at O. Then by construction OAi — OAi, and 0J5i = OB^, and A,B, = AiBi. Hence the two triangles OAiBi and OA^B^ are in all respects equal and the line AiBi may be brought to coincide with AiBi by rotation about the point O. If AiBi is parallel to AiBi , we have translation only and the point O is at an infinite distance. CHAP. II.] DISPLACEMENT OF A RIGID SYSTEM. 189 Since the angle A.OBi — the angle A-^OB^ , if we take the angle A^.OBi from both we have AiOA^ = B1OB2. If then the displace- ment is such that Ai falls on OBi or on OBi produced, Ai must be on OB 2 or OB 2 produced. In both cases OC and OD coincide and do not intersect, but it is evident that in such case the point O in which AiB, and A-.B^. inter- ^sect is the point about which rotation would produce the given dis- placement. If in any case A^A^ and B1B2 are indefinitely small, the point O is called the instantaneous centre of motion. Any Displacement of a Rigid System. — Any displacement of a rigid system may be produced by rotation about an axis and a trans- lation in the direction of that axis. Let AB and BC represent the resultant translation and rotation to which the compo- nent translations and rotations of the system can be reduced (page 187, Cor. 3). Draw AD and DB parallel and perpendic- ular to BC. Then the translation AB is re- solved into the two components AD and DB. But the resultant of DB and BC (page 187) is an equal rotation about an axis parallel to BC. Hence the ti'anslation AD and the rotation BC are reducible to the translation AD and a rotation about an axis parallel to AD. Composition and Resolution of Translation and Angular Ve- locity or Angular Acceleration. — Let a rigid system have an \ angular velocity ( ^ \ ^radians per sec. ( about an I angular acceleration ) (a radians-per-sec. per sec. ) axis AOB through the point O, and let AB — \'\''^ihe the linear rep- resentative. If we take any other point of the system, as O. , and at this point apply two equal and opposite 3 angular velocities ) a = \ angular accelerations ( ' ' ^ |i:[andO,6=j+:f,itisevi- * *"'•' dent that the motion of the system is not affected. We have then the ] angular^accelemtion [ ^^"* the axis AOB reduced to an equal j angular^ acceleratfon [ ^'^ about a parallel axis through Oi and a couple represented by AB and da. The moment of this couple is the same for every point m its plane (page 186) and equal to ] p^ [ » where p is the perpendicular distance between the two axes. But we have seen (page 177) that the moment j ^^ [ gives the linear j acceleration f\^^^ direction perpendicular to the plane of the couple. Since the moment of the couple is the same for every point in its plane, we have then translation of the entire system m a direction perpendicular to the plane of the couple, as well as simultaneous rotation about the axis through Oi. The direction a-<: 190 RIGID SYSTEM — TRANSLATION AND ROTATION. [CHAP. 11, and magnitude of this translation will depend upon the point Oi , but the rotation will be the same wherever the point Oi may be taken. Hence,* an \ ang^laraccSicm \ «/ « ^^^^ 'y'*^^ «^«^^ «^^ aads can he resolved into an equal \ angZalZcTeferaficm \ «^«^^* « parallel axis at any distance p, and a ] fjp^pj^fJ^fjfjf. [ of translation ) / ^ '^ X *w a direction at right angles to the plane of the axes. c'onvLely, m re^yltant of an j ^^"^^^Jf^T^i'Ln . i »/ " rigid system about a given axis and a \ rif^ti^?,^fi(y~, [ Of translation \j\in any direction is an equal \ ^^^-^i'Lc^^^^^^ \ «^«-* « f V 1 p = - 00 parallel axis distant \ . > ina direction perpendicular to the [ " J plane of \ ^i and the given axis. This parallel axis is the instantaneous axis. CoR. 1. Hence if we have any number of component ] anfular accelerations [ ^^*^^* ^^y axis, each can be reduced to an equal point of the system, and a \ acceleratTon [ °^ translation of the system. We can then find the resultant \ acceleration C ^^ trans- lation as on page { g [ and the resultant | a„*"uCLoetrlS,„ ( "« on page 176. Cor 2. Therefore any number of component ] o!!f!!lar ac- celSfons ( ^^d ] acletriifons \ «* translation, can all be reduced to a single resultant j onceleratwn 1 °^ translation and a single re- «^t^^* 1 aS^i^trrnSfon h^«^t ^^ ^^^ ^^^^^^^h any one point of the system. The j acceleration [ ^^ translation will vary in di- rection and magnitude with the point chosen. The ] anfuKr aopplpration [ ^^^^ ^® ^he same no matter what point is chosen. Central Axis. — Any number of component angular velocities of a rigid system can be reduced to a single angular velocity about a determinate axis and a simultaneous velocity of translation of the system along that axis. Such an axis is called the central axis, and such motion is called screw motion. * Compare Statics — Non-concurring Forces. CHAP. II.] CENTRAL AXIS. 191 Thus let OA and OC represent the resultant linear velocity of translation Vr and the resultant angu- r^ lar velocity cjr, to which, as we have just seen, all the angular velocities can be reduced. Draw AD and OD parallel and per- pendicular to OC. Then the velocity of translation OA = Vr is resolved into the two components AD and OD. But the resultant of OC and OD is an equal angular velocity about an axis parallel to OC (page 190). Hence the velocity of translation OA = Vr and the angular ve- locity OC = oor are reducible to an equal angular velocity about an axis parallel to OC and a linear velocity of translation AD along that axis This axis is called the central axis, and may be located by the following geometric construction. At any point O of the system taken arbitrarily let the velocity of translation be V,- and the ro- tation axis through O be oor, making the angle (p. Through O draw a line OD =p perpen- dicular to the plane of Vr and GOr, SO that poor = Vr siu 0, or Vr sin <p p = OJr Then a line through D parallel to the rotation axis at O will be the central axis. (Compare Statics— Non-concurring Forces. Screw Motion. — Let u,- denote the resultant velocity of trans- lation along the central axis. This is called the velocity of advance. The distance d advanced during one complete rotation of the sys- tem is called the pitch of the screw, and the distance advanced dur- ing a rotation of one radian, or —, we call the unit pitch of the screw. • ., -• £ If GOr is the resultant angular velocity of rotation, the tmie of a 2^ complete rotation is ^ = — 00 r We have then for the value of the pitch d = Vrrt = and for the unit pitch 2itUr hence w = Ord "2^' (1) (2) A— — 2it oor If r is the radius vector of any point of the system, then the linear velocity of that point due to rotation about the axis is v = roor (^) in a direction perpendicular to the plane of the central axis and the radius vector. The resultant velocity at that point is then Or = {/u,^ + ^'' = "^'•y ^ + ^ (4j. 192 EIGID SYSTEM — TRANSLATION" AKD EOTATION. [CHAP. II. The inclination of the path at that point to the plane of rotation is given by d 2nr' tan 1 = — (5) or the tangent of the angle of inclination at any point is equal to the ratio of the pitch to the circumference of the circle described by the point relatively to the axis ; or it is equal to the ratio of the unit pitch to the radius vector of the point. Centre of Parallel Angular Velocities.* — Let &n, 0^2,0,73, etc., be any number of parallel angular velocities passing through the points ^1 , A^^Aa, etc. , of a rigid system. 7"* X- Then the resultant ^r must be parallel to the components and equal in magnitude to their algebraic sum, or OOr = Gl>i + <i52 + 003 + . . . = 'Sqo. Take any two components rai and (»2, and produce the line A\ , Ai to intersection K with the plane ZX. Drop perpendiculars AiBi, AiBi to this plane and draw the line KB1B2 in this plane. Now, from page 181, the resultant of goi and G02 is g,7i = gji + (a2 and its point of application is at A on the line AiAi , so that ooi _ A-iA GOi AiA' Drop the perpendicular AB to the plane ZX. Then we have by similar triangles A,A ^ B2B A,A~B,B' Denote the distance AiBi , A2B2 by yi , y-i, respectively, and the distance AB, or the ordinate of the point of application of the re- sultant ^1 of coi and co-x, by yi. Then we have by similar triangles Hence In the same way for three angular velocities, mi , &32 , cja , we can ■combine the resultant oi of ooi and 002 passing through A, with 003. * Compare Statics — Parallel Forces. B2B _ 2/2-2/1 BrB yy-yr' CSJl _ Vl — ?/l' "^ "' <"' + ^-'^ •CHAP. II.] ROTATION AND KECTILINEAE TRANSLATION COMBINED. 193 "We thus obtain for the ordinate of the point of application of the resultant of three forces In general, then, for any number of parallel angular velocities we have for the ordinate y ot the point of apphcation of the result- ant — 2ayu »=^ « In precisely similar manner, if we denote the distances AC and AD of the point of^application of the resultant from the planes YZ and XY by x and y, we have -=^-- <^> -S (3) Equations (1), (2) and (3) give the co-ordinates of the point of application of the resultant for any number of parallel angular velocities. This point is called the centre of parallel angular veloc- ities. The same equations evidently hold for parallel angular ac- celerations, by replacing oj by a. We have then the centre of parallel angular accelerations. > The position of this centre depends only upon the magnitude and position of the line representatives and is independent of their com- raon direction. If z is zero, Zi, Zt, etc., are zero, and the line representatives all lie in the plane XY. The centre is then given by (1) and (2). If 2 and y are zero, the centre is in the axis of Xand is given by (2). Rotation and Rectilinear Translation Combined. — Let a rigid system have an angular velocity Ob = go about an axis through O, perpendicular to the plane of the paper, and at the same time a velocity of translation u in a straight line. Then, as we have seen, page 177, v can be re- placed by the couple Oa and IB, and we have at any instant a resultant rotation IB — CO about a parallel axis though I at a distance OI = p = - in a direction GO perpendicular to that of v. This axis is the instantaneous axis; that is, the point /at any distant has the velocity v in one direction d\ie to translation, and the velocity v = poa m the other direction due to the couple, and is therefore at rest. It is evident that every straight line in the system parallel to the moving axis at O and at a constant distance from O of p = - becomes in turn the instantaneous axis when it arrives at GO the position J with reference to O. Hence when a rigid system has a velocity of translation m a straight Hue and at the same time an angular velocity go about a given axis Ob, the resultant motion of the system ts the same as tf 194 EIGID SYSTEM — TRANSLATION AND EOTATION. [CHAP. lU a cylindrical surface fixed in the system, of radius p, = — , rolled on^ GO a plane HIH parallel to the plane of Oh and v. The path described by any point in the axis Ob is a straight line. The path described by any point not in this axis is called a trochoid. The special form of trochoid described by any point in the cylindrical surface is called a cycloid. Any internal point describes a prolate cycloid ; any external point, a curtate cycloid. Curtate Cycloid The general form of these curves is shown in the accompanying figures. A common illustration of such motion is a wheel rolling in a straight line on a plane. If the radius is p, we have poa—v and hence — = p. 00 The in- Cyclold stantaneous axis is at right angles to the plane of the wheel and passes through the point of contact with the plane. The velocity at the centre is t, and at the opposite extremity of the diameter through the point of contact 2» in the direction of translation. The velocity of any point at a distance d from the instantaneous axis is dco in a direction per- pendicular to the plane of the instantaneous Prolate Cycloid ^^jg ^mj i\^q instantaneous radius vector d. Combined Parallel Rotations — One Axis Fixed. — Let a rigid system rotate with the angular velocity ooi about a moviiig axis at Oi , and at the same time let this axis revolve with the angular velocity oo^ about a parallel fixed axis at Oa. Then, as we have seen, page 181, the resultant f.7r is in the plane of the components ooi and ooi, is equal in magnitude to their algebraic sum and divides the straight line joining Oi and Oi into segments inversely as the components. Also when the components act in the same direction the resultant lies within the components, and when in opposite directions without the components and on the side of the larger. Fig. 1, then, represents the case in which ooi and oai are in the same direction ; Fig. 2, that in which £»i and ao-i are in op- posite directions and coi is the greater ; Fig. 3, that in which ooi and ooa are in opposite directions and cit is the greater. The resultant angular velocity is in all cases then given by Fig. 1. GOr = GJi + ftJa, (1) where we take kji and ftjs with their proper signs. This resultant rotation oor takes place about the instantaneous axis through J, so situated that (page 181) GOi OOi 10, 10.' (2> so that at any instant 1 has two opposite and equal linear velocities and is therefore at that instant at rest. Since then ooi . 10 1 = go, . lOi, CHAP. II.] ROTATION ABOUT INTERSECTING AXES. 195 we see at once that in Fig. 2, oa, is greater than c»2, and in Fig. 3, wi is less than Wi. We have also as on page 181, taking moments about Oa and Oi , fiji . OiOi = oor . lOi, or lO-i = — . OiOa ; QOr ftJa . OiOa = COr. lOi, OT lOi = -— . OiOa. (3) All the lines in the system which successively occupy the position of the instantaneous axis are then situated in a cylindrical surface described about Oi with the radius lOi = — ^ . OiOa; and all the COr positions of the instantaneous axis are contained in a cylindrical surface described about Oa with the radius JOa = — ^ . O1O2 Hence the resultant motion of a rigid system which rotates about an axis O: while at the same time this axis revolves about a fixed axis Oa is the same as if a cylindrical surface of radius 10 1 = — ^ . OiOa , fixed in the system, rolls upon a fixed cylindrical surface oar of radius lOi = -^ . OiOa. CO,- In Fig. 1, a convex cylinder rolls on a convex cylinder; in Fig. 2, a smaller convex cylinder rolls within a larger concave cylinder; in Fig. 3, a larger concave cylinder rolls upon a smaller convex cylinder. The path described by any point in the moving axis through Oi relatively to the fixed axis at Ci is a circle. The path described by any point relatively to the fixed axis is called an epitrochoid when the rolling cylinder is outside of the fixed cylinder and an hypotrochoid when it is inside. The special form of epitrochoid or hypotrochoid described by a point in the surface of the rolling cylinder is called an epicycloid when the roll- ing cylinder is outside and an hypocycloid when it is inside the fixed cylinder. , , . ^, When the distance OiOa is infinite we have the case of the pre- ceding Article, of a cylinder rolling on a straight line. In this case When a cylinder rolls externally upon another of equal size, the special form of epicycloid described by a point in its surface is called the cardioid. In this case coi and co are equal and in the same direction. When a cylinder, as in Fig. 2, rolls withm a concave cylinder of double its radius, we have GO, — 2oo2. In this case each point in the surface of the rolling cylinder moves to and fro in a straight line, being a diameter of the fixed cylin- der; each point in the axis of the rolling cylinder describes a circle of the same radius as that cylin- der and any other point in or without the rolling cvlinder describes an ellipse of greater or less eccentricity, having its centre in the fixed axis at C. This principle has been made available in instruments for drawing and turning ellipses. Rotation about Intersecting: Axes— One Axis Fixed. — Let Ca be a fixed axis and about it let the plane O.CO, rotate with the 196 RIGID SYSTEM— TKANSLATION AifD ROTATION. [CHAP. II. angular velocity oja. Let COi be an axis in the rotating plane, and about that axis let a rigid system rotate with the angular velocity tt»i relatively to the rotating plane. If we lay off from C the line representatives Ca = ooi and CO = cBi along the axes, the diagonal Ci of the parallelogram gives the magnitude and direction of the resultant angular velocity oor . The instantaneous axis then occupies the position CI. If we denote the angles ICd and ICOi by aa and a, , we have tan aa = tan ai = GOi sin (oTi + ai) Oh + ooi cos («! + ai) sin (ai + at) — am (ai + ai) OOi 1 + — cos (ai + ai) Oil 1- cos (ai + ai) OOi (1) (2) From (1) and (2) we can find ai and a» , when the angle between the axes (ai +a ») and the angular velocity ratio — are given. We GOi have also and GOr OOi GOr' = tai" + coi^ ± 2ooicOi COS (ofi + eta), . . sin (ai + ai) oor sin (ori + aa) coi sin aa sm «i Sm rra sm a I (3) (4) All lines which come successively into the position of the in- stantaneous axis are in the surface of a cone described by the revo- lution of CI about COi ; and all the positions of the instantaneous axis lie in the surface of a cone described by the revolution of CI about COi. Hence the motion of the rigid system is such as would he pro- duced by the rolling of the cone CIOi , fixed in the system, about the fixed cone Cld. If ra is the radius dl of the fixed cone, and ri the radius Od of the rolling cone, we have rict>i = raoaa , or — Vi ri For the height Cd = hot the fixed cone we have ri + ra cos (ai + as) hi — Vi cotang era = sin (ai + as) (5) (6) CHAP.II.] ANALYTICAL DETERMINATION OF RESULTANT VELOCITY. 197 and for the height COi = hi of the rolling cone h. = r. cotang a. = r. + r. cob ja. + a,) sin (jxx + aa) (7) Pi J The plane through the instantaneous axis and the axis of the fixed cone passes through the axis of the rolling cone and turns about the axis of the fixed cone with the angular velocity ooi. The motion of this plane is called the precession and &Ja is the angular velocity of the precession, or, as it is sometimes called, the rate of precession. If a-i is zero, the fixed cone becomes a cylinder. If ai is zero, the rolling cone becomes a cylinder. If both are zero, we have the case of the preceding Article. If oit is less than 90" and ai is less than 90°, we have a convex cone rolling on a convex cone, and looking from C along the axes COs and CI the precession and rotation about the instantaneous axis are both clockwise or both counter-clockwise. This is called positive processional rotation. It is the case of a pair of bevel-gear wheels, or of a spinning top whose point is at rest. If aa is a right angle, the fixed cone becomes a flat disk with centre at C. If ai is a right angle, the roll- ing cone becomes a flat disk with centre at C. If a^ is a right angle and aa is zero, we have a cylinder rolling on a plane. If iXi is obtuse, we have a convex cone rolling inside a concave one, and looking from C along the axes COa and CI, if the precession is counter-clockwise the rotation about CI is clockwise or vice versa. This is negative precession. It is the case of the pro- cessional motion of the earth's axis. If ai is obtuse, the rolling cone becomes concave and we have a concave cone rolling on a convex cone. This is also positive pre- cession. The path described by a point relatively to the fixed axis is called a spherical epitrochoid or hypotrochoid according as the rolling cone is outside or inside of the fixed cone. The special form of spherical epitrochoid or • l'r.-o.li hypotrochoid described by a point in the sur- face of the rolling cone is a spherical epicycloid or hypocycloid. Analytical Determination of Resultant Angular Velocity and Velocity of Translation for a Rigid System with Any Number of Non-concurring Angular Velocities.* — (Compare Statics — Non- concurring Forces.) Take any point O of the rigid system as the origin of a system of rectangular co-ordinates. Let the com- ponent angular velocities be ca,, c^-2, etc., making the angles («., /Si, rx\ (aa, /ia, r^), etc., w-ith the axes of X, F, Z, respectively. * Angular accelerations are treated in precisely the same way as angular velocities, and every equation in this Article can have oo replaced by a. The student should make such substitution and interpret the results. 198 RIGID SYSTEM — TRANSLATION AND KOTATION. [CHAP. IL, Resultant Angular Velocity. — We have, just as on page 65, re- placing V by cj, for the component angular velocities parallel to X, Y, Z, eox = <»i cos ai + oOi COS a^ + . . . = 2ol)COS a ; 1 a>l, = ooi COS /ii + ooi COS pi + . . . =. 20JC0S /H; J- • • • • (1) GOz = ooi COS Xi + oOi COS /a + . . . = ^&>COS y. J The resultant angular velocity is ojr = |/ffi>a;* + aJj/" + t»/, . (2) and its direction cosines are cosa= — , cose=-^, cos/=-— (3; COt OOt GOf The magnitude and direction of the resultant angular velocity are thus determined. Resultant Velocity of Translation. — Let (Xi, yi, Zi), {Xi, y^, Zj), etc., be the co-ordinates of points on the rotation axes of wi, gljj. etc. We can resolve each angular velocity oji, cou, etc. (page 190j, into an equal angular velocity about a parallel axis through the origin O, and a velocity of translation of the system due to a couple, given by the moment of «»j, <»2, etc., with reference to O. We can thus re- duce the given angular velocities to a resultant angular velocity o^,- about an axis through the origin O, whose magnitude and direction are given by (1), (2) and (3), and a velocity of translation v,- of the axis through O. The components Vx , Vy , Vz of this velocity of the axis through O, along the axes of X, Y, Z, are therefore given by (compare Statics — Non-concurring Forces) Mx =Vx= 2a)y cos X — 2a)Z COS /S ; ^ My = Vy= 2a)Z COS a — 2oox COS y; > (4) Mz = Vz = 2oox COS /3 — 2ooy cos a. J For any other point P whose co-ordinates are {x', y", z'} we have simply to put x — x', y — y', z — z' in place of x, y, z in (4) and we have for the components of the velocity of this point along the axes Vx = 2a}y COS y — 2ooz COS /3 + {ooyZ' — oozy') ; ] Vy — 2ooZ COS a — ^caa? COS y + {a)zX' — gJxZ') ; I • • • (5) Vz = SooXCOS /3 — 2ooy COS a -\- {ooxy' — ooyX). J Let us write Vx = a>yZ' — oozy' ; 1 Vy = OOzX' — GOxZ' ; l (6) Vz = OOxy' — GOyX'. J Then we can write in general for the components of the linear velocity of any point whose co-ordinates are {x\ y', z') Vx = Vx + Vx ; 1 Vy=Vy + Vy'\ 1 (7) Vz=Vz + Vz; J where Vx,Vy^ Vz are given by (4) and Vx', Vy', Vz by (6). CHAP. II.] CONDITIOlf FOR A SIlfGLE AKGULAK VELOCITY. 199 If the resultant axis of rotation passes through the origin O we have vx = uj, = 0, vz = 0. Therefore equations (6) give the com- ponents of the hnear velocity of the point P due to rotation about an axis through the origin O paraUel to the resultant axis ihe resultant linear velocity for any point is then in general Vr=\/Vx'+Vy'+Vz'; (8) and its direction-cosines are cosa=^, co86 = ^, cosc = ^ (9) The magnitude and direction of the resultant linear velocity of any point are thus determined. Conditions of Rest.*— If the system is at rest we must have, Ist, Vx, Vy, Vz equal to zero, or, from (7), Vx = 0, v„ = 0, Uz = 0; and also, 2d, v'x = 0, V y = 0, v'z = 0. We see from (6) that the second condition is fulfilled when a)x=0, coy = 0, ajz = 0, that is, when oor = or there is no rotation. In this case all the angular velocities must reduce to two equal and opposite resultant angular velocities. The first condition is fulfilled when equations (4) are zero. That is, the two equal and opposite resultant angular velocities must pass through the same point, so that their moment is zero at any point. We have then for the equations of condition of rest, from (1), 2aoe08a = 0; ] 2&jCOS/S = 0; I (10) Saocosy = 0; J and, from (4) Saoy cos r — ^ooz cos /? = 0; "j SooZCOaa — Saoxcosy = 0; I (H) 2oox COS /3 — 2ooy cos a = 0. J If equations (11) only are fulfilled, then the two opposite resultant angular velocities pass through the origin, which is therefore at rest; but unless (10) is also fulfilled they are not equal, and we have rotation about an axis through O, but no translation. If equations (10) only are fulfilled, there is no rotation, the two resultant angular velocities are opposite and equal, but unless (11) is also fulfilled they do not pass through the same point. Hence they form a couple, and we have translation and no rotation. Condition that the Angular Velocities shall Reduce to a Single Angular Velocity. — If the angular velocities, then, all intersect in one point of the system, the moment at that point is zero. It has therefore no translation and the system rotates about an . axis through that point. If the angular velocities do not intersect in a single point, we have in general translation and rotation. There is, however, one case in which the angular velocities may not all intersect in one point, and yet we may have rotation only without translation. In this case the angular velocities must re- duce to three, any ttco of which intersect, while the other, although it does notpdss through ilieir point of intersection, yet intersects their resultant. * Compare Statics — Non-concurring Forces. 200 RIGID SYSTEM— ^TBANSLATIOK AND KOTATION. [CHAP. II. Thus let the resultant angular velocities gox, coy intersect in a point A. We can then take them as acting at any point in their resultant AC. Let ooz intersect AC at S. Then we can take all three acting at JB, and we thus have rotation only, about an axis through_5^ _ Let ar, y, z be the co-ordi- nates of B. Then, since we can take oox, a>y, ooz at B, we have for the components of the velocity of the origin O , (12)' My Mx = Vx = oozy — a>yZ ; ^y = Vy = OOxZ — OOzX'i Mz = Vz =^ OOyX — ooxy. If we multiply the first of these by oox , the second by coy , and the third by wz and add, we have (compare Statics — Non-concurring Forces) Vx OOx + Vji QOy + Vz OOz =^ (13) We should obtain the same result for any other two components intersecting and a third passing through a point on their resultant. Equation (13j then gives the condition that all the angular ve- locities acting upon the system reduce to a single angular velocity at a point whose co-ordinates are x, y and z, and we have rotation only (page 179). We have evidently for the equations of the projection of the line of the resultant on the co-ordinate planes GOu Vz OOx Vv GOz Vx y = -!Lx ~ , X = Z — ^-, Z ^ ^^y . <ax GOx OOz GOz OOy OOy Parallel Velocities.^ — (Compare Statics — Non-concurring Forces.) If the axes of all the angular velocities are parallel, we have «, /^, y constant and the same for all. Hence from (3) and (1) OOx = oor COS d = cos a'Soo ; 1 Wy = oor COS e = COS /32a3 ; [► (14) OOz — GOr COS / = COS y2oo. The resultant oor must have the common direction of the paral- lel components, or d = a, e — (i, f = r, and ca^ = ^oo ,-.... (15) that is, the resultant angular velocity is equal to the algebraic sum of all the parallel components and is parallel to them. For any point of the system whose co-ordinates are x', y', z', we have from (4), by putting {x — x'}, (y — y'), (z — z') in place of x, t/, z and taking a, (i, y as constant, V^ — cos -{tioiy — y') - cos fi'S.uii.z - z') = cos 7[2a>// - ?/'2(o] - cos /SrSux - z'Soj] ; \ F„ = cos a2<u(Z - Z')- COS y^S.u,(x - x') = cos a[2a>Z - z'2a>] - COS ■) [2wx- - a:'2<o] ; y . (16^ Fg = COB P2«(«— »•) - COS aXmiy - y') = COS (3[2wa; - a:'2w] - cos a[2w^- j/'2w]. ) CHAP. II.] COMPONENTS OF MOTION OF A RIGID SYSTEM. 201 If we substitute (16) and (14) in (13), we see that equation (13) is. satisfied. We have therefore a single resultant velocity and rota- tion of the system about a fixed point. This point is given by the values of x\ y', z' which make Vx, Vy, Vz zero. The point there- fore whose co-ordinates are — 2(ax — 2coy — 2a)Z *=^^' 2/=-^. ^=-2^' (17) is at rest and the resultant axis must pass through it. This point is called the centre of parallel velocities (page 193). Any other point has a velocity given by (16). If 2 go = 0, the resultant axis is at an infinite distance, or there is no rotation, but translation only, given by (16). Components of Motion of a Rigid System.— In order that the motion of a rigid system at any instant may be known, it is sufli- cient to know the velocity at that instant of some given point of the system, and the rotation of the system at that instant. Take the given point always as the origin. Then the velocity of this point is known when its components Vx, Vy, Vz along the axes are given, and the rotation is known when the components Gox, ooy, Goz of the augular velocity along the axes are given. The motion of the system at any instant is then known when these six quantities, Vx, Vy, Vz , cox, ooy, ooz are given. These six quantities are called the components of motion of the system. Equivalent Screw. — (Compare Statics — Non-concurring Forces.) The motion of a rigid system being thus known, it is required to find the screw motion to which it is equivalent. That is, to find the central axis, the linear velocity along the central axis, and the angular velocity about it. Since Vx, Vy, Vz, oox, wy, aoz are given, we have: (1) The angxilar velocity about the central axis GOr = Vo^x' + oOy'' + '^z. (1) (2) The direction cosines of the central axis cos d = -^, cos e = — ^, cos/ = — (2) ca,. 00^ oor (3) The linear velocity of every point resolved in a direction parallel to the central axis must be the same and equal to that along the central axis. Let u,- be the resultant linear velocity of every point of the system along the central axis, and let its compo- nents along the co-ordinates axes be Ux, Uy, Uz- Take the point for which Vx, Vy, Vz are given, as the origin, and let the co-ordinates of any point of the central axis be xf', y", z". Then the components Vx, Vy, Vz of the velocity of the origin due to rotation about the central axis are, from equations (1), page 179, Vx= oDztf" — f^yz"; Vy = ooxZ" — o^zX"; ■ (3) Vz = ooyx" — o^xV". We have then Vx=Ux+ Vx, Vy = Uy + Vy, Vz = Uz + Vz i or llx= Vx — Vx, 'iMl=Vy—Vy, Uz=Vz-Vz.. . . (4). 202 EIGID SYSTEM — TRAlfSLATIOK AND ROTATIOlf. [CHAf*. II. Hence Mr = {Vx — Vx) COS d + iVy— Vy) COS 6 + (Vz — Vz) COS/. (5) Inserting the values of the direction-cosines of the central axis from (2), we obtain Ur a)r={Vx— Vx)0Ox + (Vy — Vy )0Oy + (Vz — Vz)(^z. But since Vx^x + Vy^y + Vz ooz = 0, this becomes Vn- oa,- = VxO!>X + Vy (>^y + Vzl^Z (6) We also have from (4) Mr COS d=Ux=Vx— Vxi «*r COS 6 = Vy — Vy , Ur COS f = Vz — Vz. (7) Hence from (2) and (3) Ur Vx+ f^yZ'' — oozy ' _Vy + oOzX" — oo^z" Vz + ^xV" — <^yX" ._. = ■ — — . ( o » cOr oox ooy ooz Equations (8) give the equation of the central axis. From 1 6 ) and (1) we have Ur^ _ Vx^X + VyCOy + Vz^Z ^gx This we have called the unit pitch (page 191), or the distance of advance during a rotation of one radian about the central axis. If we substitute (9) in (8j and reduce, we have for the equation of the central axis 1 / ,, VzOOy — Vyooz \ _ 1 / ,, VxOOz — VzGOx \ oox \ <»r' l~ <^y\ <»'•' I . ^±L._ Vy^x-VxC^y \ ^ (^0^ Therefore the central axis passes through a point whose co- ordinates are * _ FgCJy — VyC^z „ _ Vxf^z — VzCOx » _ VyOOx — VxOOy .^^. If we substitute these values of x", y", z", in (3) and (7), we have from (2) Vx = u,- cos d — <»r(2;" COS e — yi' cos/), <»« = <»>• cos d\ 1 Vy = Ur COS 6 — oa,(a^' COS / — z" COS d), OOy = OOr COS 6; |- . (12) Vz^'Ur COS / — ^r(y" COS d — ic" COS e), <»2 = ca,- COS/. When, therefore, the components of motion Vx, Vy, Vz, f^x, coy, ooz are given for any point, we find <», from (1), the direction of the central axis from (2), and the position of the central axis with reference to that point as an origin from (11). We have also the velocity of advance Ur from (9). * Since velocity in the liodograph is normal acceleration in the path (page 52), Vzooy — Vyooz is the component in the direction of Xoi the normal linear acceleration of the origin due to rotation about the central axis. The normal Vzo^v — VyOOz linear acceleration is poor^. Hence — ^— ^ is the projection of p on the «xis of X. CHAP. II.] COMPOSITION AND RESOLUTION OF SCEBWS. 203 On the other hand, if the position of the central axis 'x" v" z") IS known together with the linear velocity Ur along it and the 'an- gular velocity oor round it, the components of the motion for the origin are given by (12). The Invariant.— (Compare Statics— Non-concurring Forces) From (6) we see that the quantity VxOfx + VyGi)j/ + VzGOz is always equal to m, to, , and is therefore invariable no matter what point is taken and whatever the values of a>x, ooy, aoz, that is, what- ever the direction of the axes. This quantity is therefore' called the invariant of the components. Since oo,. is also invariable what- ever point is taken and whatever the direction of the axes, it may be called the invariant of the rotation. If the motion is such that the invariant is zero, it follows that either m,- = or ca,- = 0. The condition VxOOx + VyCOy + VzOOz = is therefore the condition that the motion is equivalent to either a simple translation or a simple rotation. If &?)• is not zero and this condition is fulfilled, we have rotation only (see pages 179, 200). Composition and Resolution of Screws. — (Compare Statics— Non-concurring Forces.) If two screws are given, then by equa- tion f 12) we can find the six components of motion of each screw. Adding these two and two, we have the six components of the resultant screw. Then by (1), (2), (6) and (11) the central axis together with the linear and angular velocities of the screw may be lound. Conversely, we may resolve any given screw motion into two screws in an infinite number of ways. Since a screw motion is represented by six components at any point, we have in the two screws twelve quantities at our disposal. Six of these are required to make the two screws equivalent to the given screw. We may therefore in general satisfy six other conditions at pleasure. Thus we may choose the axis of one screw to be any given straight line we please with any given linear velocity along it and any angu- lar velocity round it. The other screw may then be found by re- versing this assumed screw and joining it thus changed to the given motion. The screw equivalent to this compound motion is the second screw, and it may be found in the inanner just explained. Or again, we may represent the motion by two screws whose unit pitches are both zero, the axis of one being arbitrary. We can thus represent any motion by two angular velocities, one, <», about an axis which we may choose at pleasure, and the other, w', about some axis which does not in general cut the first axis. These are called conjugate axes. These angular velocities are such that air would be their resultant- if their axes were placed parallel to their actual posi- ^ tions, so as to intersect the central axis. If then d is the ^\i^ shortest distance between the axes, we have Fr = dcj; and if ^ is the angle between &? and «r , and the angle between co and oo, we have Also Ur Vr sin ^ = ttr , or smip = ==-. Vr . . . , fij' sin 9 fijr sin if} = 00 sm 9, or sin ip = cor 204 RIGID SYSTEM — TRANSLATION AND ROTATION. [CHAP. 11^ Hence "Fi-fij'sine =tlr00r, or doooo'ainB = Uroor', hence UrCOr d = Gooa' sin 9 (1> EXAMPLES. (1) A line DE moves, keeping its extremities in tico fixed lines ADB, AEC. Find the instantaneous centre and the direction of motion of any point G at any instant. Ans. From D and E draw BF and EF perpendicular to AB and A C, meet- ing at P. Then F' is the instantaneous centre (page 189). Join OF. The di- rection of motion of G is perpendicular to OF. (2) A line DE moves, keeping its extremities in two fixed lines, one, ADB, vertical and the other, AEC, horizontal, and makes at a given instant an angle of 30° with the horizontal. Find (a) the di- rection of motion of the middle point of DE at the instant, and (b) the point whose motion is inclined at that instant 30° to AC. Ans. (a) Inclined 60° to ^C ; (b) ~DE from E. (3) A line moves so that its extremities remain in a given circle. Find the instantaneous centre of motion for any instant. Ans. The centre of the circle. (4) Find the ratio of the velocity of any point of a screw to its velocity of advance. / d where d is the pitch, r the radius of the screw (page Ans, 191). (5) Let Ci and d he fixed axes about which turn the cranks CA\, CiB, whose free ends are connected by the link AB, jointed at A and B. The axes are perpendicular and the plane of motion parallel ta the paper. If the linear and angidar velocities of A are Vi, ooi, find the linear and angular velocities Vt , wi of B. Ans. Let ACi=ri and BCt — r^. Produce CiA and dB to meet in J. Then at any instant the linear velocities of A and B are perpendicular to A C» and BGi respectively. Hence at that instant AB is rotating about the instan- taneous axis at /. Let co be the angular velocity of AB about /. From d , Ca, /let fall the perpendiculars C\D, VtF, IE on the line of the link AB or CHAP. II.] EXAMPLES. 205 its prolongation. Also draw the line of centres Ci C, cutting the link, pro- longed, if necessary, in the point K. Then TidJi = AI . 00, or — = rjftJa = BI. CO, or also, — Al_ IE CD' BI _ IE C,F C,K Since naji = Vi and r^oo, = tJa , we have Ti _BI v,~ AI _ Hence — 1. The linear velocities of B and A are to each other as their distances from the instantaneous axis. 2. The angular velocities of the cranks are to each other inversely as the perpendiculars from their centres of motion upon the line of the link ; or in- versely as the segments into which the line of the link cuts the line of centres. (6) In the case of the crank and connecting rod, since B moves in a straight line CiB, we have BI always perpen- <licular to GiB, and hence «i BI AI' or Vi = AI TiGOi. Let the distance CiB = s, the length of the connecting-rod = I, and the angle of the crank r, with CiB = 61. Then we have BI= 8 tan I AI = cos 01 - n, s = ri cos 6, + Vi* - ri^ sin* 6,; or, if I is very long compared to ri , approximately ri* sin* fli >.t?. 8 = Ti cos 61 -\- I — 21 Hence «j = Stan 01 . TiOOi « sin 01 cos 01 — Ti 8 — r, cos Bi -rifiJi. When 01 = 90, we have v, = rjcoi = «i , or the velocity of ^ and 5 are equal and B/and Alure infinite. When 0i = or 180°. we have e, = and s = l-\-ri or I— r,. These are the " dead points " of the crank, or the ends of the stroke. (7) A rod (length - I) hangs by a small ring at its upper end from a fixed horizontal rod. To the former an angular velocity go is given in a vertical plane through th^ fixed rod, so that the centre of the movable rod moves vertically. Find the linear velocity of its centre when its inclination to the vesical is 0. Ans. GolsinO. (8) A disk (radius = r) rolls without sliding on a plane. Find the relation between its angular velocity <» and the linear velocity v of its c&Yvtv^t Ans. The point of contact with the plane is at rest at any instant, or ra> = — ». 206 RIGID SYSTEM — TRANSLATION AND ROTATION. [CHAP. II. (9) A rod AB (length = J) rotates about a hinge at A and rests with its end B on the surface of a ivedge BCD. The wedge advances towards A xvith velocity v. The angles BAG = 6. BCD = (p. Find the angular velocity oo of the rod. V sin <p Ans. I cos (0 — 0)' flO) Two bevel-gear wheels have the angle between their axes 70°. The rolling wheel is required to make 3i revolutions about its axis while going around the axis of the fixed ivheel once. Find the angles of the bevel. If the inner radius of the fixed ivheel is 50 inches, find that of the rolling wheel and the length of the axes. Ans. (See page 196.) The angular-velocity ratio <Bl 7 „ — = - . Hence GO, 2 7 sin 70° _-o . ^, tang or, = » , r,„„„; ^o, or a, = 56 15', and Lence a, 2 + 7 cos 70 13° 45'. inches, and hi = 100 7 We have also ri = -=- ^ -I- 50 cos 70° = 33.4 inches, sin 70 50 + cos70° A. sin 70° = 58.4 inches. (11) The angle between the plane of the earth''s equator and the plane of the ecliptic is 23° 27' 28". The earth rotates about its polar axis in one sidereal day and makes a revolution about the axis per- pendicular to the plane of the ecliptic in 25868 years. Finathe instantaneous axis. Ans. (Page 196.) We have fij, = 27r radians per day, and co^ = ^„ „ — g^rr radians per day. Also the angle O^COi = ai-{- a Oil = r tan ai 25868 X 865i 23° 27' 28". Therefore 0,/ is r sin (flTi + aa) OOi -\- cos (ai -f aj) where r is the polar radius of the earth, or 3950 miles. The radius of the roll- ing cone is then Oil— 5.52 ft. and the angle a, = 0". 00867. (13) A rigid system has an angular velocity of ooi = 40 radians per sec. about an axis parallel to the axis of X, passing through a point whose co-ordinates are Xi = 2 ft., yi = 3ft., Zi = 0. and a simultaneous angular velocity of oo^ = 30 radians per sec. about an axis coinciding with the axis of Y. Find the resultant angular velocity and the instantaneous axis. Ans. (Page 198.) We have ftjj- = -f- 40, ajj/zr-j-SO, ea^ = 0, <», = 50 ra- dians per sec. The instantaneous axis makes the angles with the axes given by cos o ^ — , 50 30 cose = ~, cos/: 0; d = 36° 52', e = 53° 7', /= 0°. CHAP. II.] EXAMPLES. 207 Uz •*-GOx The velocity of translation of the system is given by ux =0, uy = 0, uz = — 120 ft. per sec. a)|] The condition uxoox + Uyooy -\- uzooz = is sat- isfied. Therefore there is rotation only about the instantaneous axis which passes through the intersection /of oox and cay. The velocity of any point whose co-ordinates are x = 2, i/' = 2, 2' = 3 is given by Vx — ooyz' = -|- 90 ft. per sec. ; Yy= — oox!^ — — 120 ft. per sec. ; Vz = Uz — ooyx' 4- o^a^ = — 100 ft. per sec. The resultant velocity of this point is then Vr = 180 ft. per sec. and its direction cosines are 90 ^ 120 100 cos a = rr^x. COS = tt^^, COS C = ,- 120 cos := . 180' 180' " 180' ^"""180' or a = 60°, 6 = 131° 48', c = 123° 45'. (13) A rigid system has the angular velocities ttj, = 50, ttJj = 30, (»3 = 70, cot = 90, and ooi = 120 radians per sec. about axes passing through points of the rigid system given by Xi=-h 5 ft., 2^1 =+10 ft.; «,=+ 9 ft., 2^j = +12ft,; a;3 = + 17ft.,2/3 = + 14ft.; «« = + 20 ft., y* = + 13 ft. ; a;^ = -f 15 ft., ys = + 8 ft.; and making with the co-ordinate axes the angles a. = 70°./?!= 20°; a, = 60°, /?, = 150°; or, = 120°, /Jj = 0°; a, = 150°, /34 = 120°; a, = 90°, /S. = 0°. Find the resultant, etc. (Compare Vol. II, Statics.) Ans. (Page 198.) We have for the components of the angular velocities parallel to the axes cox = 50 cos 70° + 30 cos 60° — 70 cos 60° — 90 cos 30° = — 80.842 rad. per sec. ; ooy = 50 cos 20° - 30 cos 30° + 120 + 70 cos 30° - 90 cos 60° = + 156.626 rad. [per sec. ; coz = 0. The resultant angular velocity is given in magnitude by oor = '^oox* + ooy^ = + 176.259 radians per sec, and its direction-cosines by cox - 80.843 cos d = = oor OOy 176.259 ' + 156.626 or <2 = 117° 18' 1"; or e= 27°18'1". ""° " ~ Or 176.259 ' We have from equation (4), page 198, per sec. ; , per sec "208 RIGID SYSTEM— TRANSLATION AKD ROTATION. [CHAP. II. We have then for the components of the linear velocity of the origin Vx = 0, vy = 0, Vz = 2 oox cos /3 — 2a)ycx>s a = -\-d083.dl5 ft. -per sec. Since then equation (13), page 200, VxOOx -f- i>y'*'y -\- '>^z^z = is satisfied, the angular velocities reduce to a single resultant angular velocity -and we have rotation only. The moment of this resultant angular velocity relative to the origin gives us «}• = \/vx' -\- %* + Bz* = «2 = + 3083.915 ft. per sec. Its lever-arm is vr 3083.915 ,„^^^ r = - = ,-„ q.„ = 17.5 ft. oor 176. 3o9 The equation of the line of direction of the resultant angular velocity is y = ^x - ^ = - 1.95a; + 38.14 OOx COx The co-ordinates of the point through which this resultant angular velocity passes are given from equations (17), page 201: X = ?^^2iA = + I2i ft.; ^ = 2^yco.a ^ ^ ^^^^^ COy OOx (14) Find the resultant angular velocity for a number of parallel ■angular velocities given by ooi = -\- 83 rad. per sec; «! = + 25 ft.; jti = -f 13 ft.; (»!, = + 20 " " " X2 = — 10 " Pi = — 15 " G03 = - 35 " " " X, = -f 15 " y» = - 37 " (B4 = - 72 " " " 0:4 = - 31 '• y4 = + 17 " ttjs = -f 120 " " " ais = -f 23 " Vi = — 19 " Ans. ttJr = -|- 66 radians per sec; x — -[- 77.15 ft.; ^ = — 36.82 ft. (15) Find the resultant, etc., for the angular velocities given by GOi — 50, 002 = 70, oiJs = 90, (»4 = 120 radians per sec. «, = 60° a, = 65° «, = 70° at = 75° /Si = 40°; Xi acute; Xi =0; yi = 0; a, = 0; /3j = 45°; Xi acute; Xa = + 1 ft. ; yj = -|- 4 ft. ; «» = + 7 ft. ; /Sa = 50": rs acute; Xs = +2 " 2/3 = -f 5 " Zs = + 8 " /J4 = 55°; ^^4 obtuse; X4 = +8 " ^4 = + 6 " a4 = -j-9" (Compare Vol. II, Statics.) Ans. We find the angles y by the formula, page 12, cos* y = — cos (a -|- /3) cos (a — /3). Then from page 198 we have fiJx = -f 116.423, ooy= -{- 214.480, ooz = - 51.057 rad. per sec. Therefore the resultant angular velocity is oor = i/oox^ + fi>i/* -|- 02* = + 349.325 rad. per sec, and its direction-cosines are given by cos a = , cos e = , cos/ = ; GO)- oor G^r or a - 63° 9' 48", e = 30° 39' 20", / = 101° 49'. CHAP. II.] EXAMPLES. 209 We also Lave for the components of the velocity of the origin, from equa- tions (4), page 198, vx= - 1838.604, % = -f 928.947. Tz = - 86.903 ft. per sec. The resultant linear velocity of the origin is then vr = 4/cx* + V + »«* = + 2061.789 ft. per sec., and its direction-cosines are given by Vx , Vy Vz cos a = — , cos = -=-, cos c = — ; Vr Vr Vr at a = 153° 5' 40", 6 = 63° 14' 15", c = 92° 24' 56". The equations of the projections of the resultant angular velocity on the co-ordinate planes are y = 1.885a; -f 0.746, x = - 2.28z + 18.19, s = - 0.238y - 8.57. We see that vxoox-\-i}yooy-\- Vzooz does not, in this case, equal zero. Hence ^page 179) the angular velocities do not reduce to a single resultant, but to a resultant angular velocity about the central axis and translation along that axis due to an angular- velocity couple. The resultant angular velocity about the central axis is, as already found, cor=^-{- 249.325 rad. per sec, and its angles d, e, /with the axes are already found. The co-ordinates of the central axis are given by equations (11), page 202 : ^, ^ OayVz-OO^y ^ ^^g ^ OO^x - OOxVz ^ _^ j g^g ^^ , aof* ' oor a" = '^^y-'^y^^ = . 8.08 ft. car* The resultant linear velocity Ur along the central axis is given by equation (6), page 202: GOr Its direction-cosines are the same as for oor. The components of Ur are given by equations (7), page 202 : ux = Ur cos d — — 19.481, % = «r cos e = — 35,806, Uz — Ur cos/ = -f- 8.5238 ft. per sec. (16) In the preceding example find ivhat the co-ordinates ic* , «« , z* of the angular velocity oot must be in order that all the angular velocities may reduce to a single resultant angular velocity. (Com- pare Vol. II, Statics.) Ans. We must evidently have oox, ooy , aoz , oor and the angles d, e, f un- changed, since changing the co-ordinates x^, y^, z* without changing the mag- nitude or direction of oo^ has no effect on the magnitade or direction of the resultant oor. We have then vx=- 659.571 - 93.262^4 - 68.829s4; ) «j,= -f 369.629 + 31.05924 +93.262*4; )• (1) vz= - 107.036 + 68.82fti!« - 31.0592^4. ) "We have as the equation of condition for a single resultant VxOOx + VyOOy + VzODz = 0, or 116.423Ba; + 214.48PJ/ - 51.057*2 = 0. ■Pa; + 1.842ci,-0.4386oz= (2) 210 EIGID SYSTEM— TRANSLATIOlf AND ROTATIOlf. [CHAP. !!► From (1) we obtain (Vx + 659.571) 31.059 + {vy - 369.639) 68.829 = (b2 + 107.036) 93.262, or ®a;+ 2.216% -3.003»z = + 481.034. (3> From (2) and (3) we obtain 0.374% - 2.564«z = + 481.034. If we retain for vy its value in the preceding example, -}- 928.947 ft. per sec, we shall have vz=:- 52.108, 'Ox=- 1733.975 ft. per sec. If we substitute these values in (1), we obtain 93.262^4 + 68.829S4 = + 1074.4; 31.05924 4- 93.262a;4 = + 559.308; 68.829a!4 - 31.059^4 = + 54.934. Hence a!4 = - 0.333S4 + 5.997, y* = - 0.738z4 + 11.520. If then we assume z^ = 0, we have x* = -\- 5.997, i/t = -}- 11.52 ft. (17) Using the values of the preceding example, find the point through which the resultant angular velocity passes. (Compare' Vol. II, Statics.) Ans. We have Ox = + 116.433, coy = + 214.480, coz = - 51.057, oor = + 249.325 radians per sec. ; d= 62° 9' 48", e = 30° 39' 20 ', / = 101° 49' ; i>x=- 1733.975, % = + 928.947, vz = - 53.108, Vr = + 1967.833 ft. per sec. ; a = 151° 47', b = 61° 49' 53", c = 91° 31' 3". The co-ordinates x, y, z are given (page 300) by - 1733.975 = 002^ - GJijZ = - 51.057y - 214.480sl 4- 928.947 = co:^ - oo^ = + 116.423s -f 51.057«; - 52.108 ^ooyX-a}j^= + 214.480» - 116.423^. Hence we obtain x= - 2.2802s + 18.194; y = -4.2008s + 33.961. If we assume i = 0, we have « = + 18.194, p = + 33.961 ft. If we should introduce then a fifth angular velocity, 0!J5= + 249.325, whose direction makes with the axes the angles as = 117° 50' 12", /Ji = 149° 20' 40", y^ = 78° 11', passing through a point whose co-ordinates are ajj = + 18.194, j'6 = -f 33.961 and Zi = 0, the conditions for rest (page 199) would be satisfied, and we should have QOr = 0, Vr = 0. (18) A point of a rigid system rotates about an axis at a distance of 5 feet. The linear displacement of the point is 8 ft. Find the angular displacement ana the direction of the linear displacement. Ans. sin— = — — ~- 5 = angular displacement = 106' 14' = 1.858 ra- 2 2r 5 dians. The linear displacement makes an angle of 36° 53' with the radius of rotation. -V,-+3t CHAP. II.] EXAMPLES. 211 (19) A point of a rigid system has at any instant the component linear velocities Vx= +Q,Vy= - 18, Vz = + 40 ft. per sec., and the system at the same instant rotates about an axis perpendicular to the plarte of XY with an angular velocity of 6 radians per sec. in the direction from X towards Y. Find the equivalent screw motion of the system. '' Alls. (Page 201.) We take the given point as the origin. Since the axis is perpendicular to the plane of XY, we have eaa; = 0, fij» = 0, 02 = (Br = + 6 radians per sec. Since the condition Fa;<»a;-}- VyGOy-\- Vzwz =0 is not fulfilled, we have rotation and translation combined, or screw motion. From equation (2) we have cos d = 0, cose = 0, cos/=l, or the central axis is parallel to the axis of Z. The position of the central axis is from equation (11) given by i:" = -h3ft., y" = + lft., 2"=0. It is therefore at /as shown in the figure ^ with respect to the given point 0. Substituting these values in equation (3) ^i we have for the components along the co-ordinate axes of the velocity of due to rotation about the central axis «« = + 6, % = — 18 ft. per sec., Vz = 0. Therefore from (4) we have for the components along the axes of the trans- lation of 0, ux = 0, uy = 0, uz = + 40 ft. per sec. The system, therefore, at the instant in question rotates about an axis through /perpendicular to the plane of XY, in a direction from X towards Y, with the angular velocity of 6 radians per sec. , and at the same time moves along this axis in the direction OZ with a velocity of translation of 40 ft. per sec. If the axis of rotation and angular velocity do not change in direction or magnitude, the system advances along the central axis during a rotation of one radian, a distance equal to the unit pitch, given by equation (9), viz., 6J ft. In one complete rotation then it advances a distance of 27r X 6f = 20.9 ft. This is the fitch of the screw motion (page 191). The velocity at any point, as Pi or Pa , due to rotation about the central axis is equal to /Pi . ooz or IP^ . ooz , where /P or /Pa is the radius vector or per- pendicular from the point upon the central axis. If then we take Pi as origin and the co-ordinates of / are a; = -f- 2 f t. , y = -|- 5 ft. , z = + 3 f t. , we have from (3), for the components of the velocity of Pi due to rotation about the central axis, ©a; = -|- 30 ft. per sec., ©j/ = — 12 ft. per sec, Vz — O; and since «« = 0, «j/ = 0, «« = +40 ft. per sec, the components of the total velocity of Pi are, from (4), Fx = + 30 ft. per sec, Fj/ = - 12 ft. per sec, Fa = + 40 ft. per sec In the same way for the point Ps , if we take it as origin and the co-ordi- nates of /are a; = 0, 2/ = + 5 ft., s = 0, we have Da; = +30 ft. per sec, % = 0, Vz = 0; Fc = + 30 ft. per sec, Vy = 0, Fa = -f 40 ft. per seci If the velocities Fc and Fs of the point Pa do not change in direction or magnitude, we have the case of a system translated in the direction 0? and rotating about an axis through Pa , while at the same time this axis has a ve- locity of translation in a straight line (page 193). 212 BIGID SYSTEM — TRAIfSLATION AND ROTATION. [CHAP. II. The motion of the system would then be the same as if a cylindrical surface of radius P.il= 5 ft., fixed to the system with its axis passing through Pj at right angles to the plane of XT, rolled on the plane HIH parallel to the plane Xywith the angular velocity goz — -^Q radians per sec, while at- the same time the cylinder is translated parallel to OZ with the velocity Fz = -(- 40 ft. per sec. (20) A base-ball rotates about an axis through its centre in a horizontal plane with an angular velocity ooz = — 60 radians per sec, and its centre has a horizontal velocity of translation of 1^= 50 ft. per sec. in a direction making an angle of 36° 52' with the axis of rotation. Find the motion of the ball. Ans. Let the plane of ZX be horizontal and take the centre as origin. Then, since V is in this plane, we have Fx = + 30ft. persec, Vy = 0, Fz = -|- 40 ft. per sec. Also, cox = 0, GJj/ = 0, ooz= — 60 rad. per sec. The rotation is then clockwise, or from Y towards X, as shown in the figure. Then, just as in the preceding example, the central axis is parallel to the axis of Z, and the position of the central axis is, from equation (11), page 203, given by x" = Q, y' =01= -l^it., 2" = 0. If then we neglect the acceleration due to the attraction of the earth, the motion of the ball is a screw motion consisting of a velocity uz = + 40 ft. per sec. along the axis of rotation OZ through the centre of the biill, and a rota- tion of caz = — 60 radians per sec. about this axis, together with a translation of this axis of Vx = +40 ft. per sec. Or, neglecting the acceleration due to gravity, the motion is the same (page 193) as if the ball were part of a cylinder of radius 01= -^ ft. whose axis OZ a is the axis of rotation of the ball, and this cylinder rolls on the horizontal plane HIH with angular velocity gjz = — 60 radians per sec, while at the same time the cylinder is translated along OZ with the velocity -f- 40 ft. per sec The centre of the ball moves then in the resultant of V- and Vz , or along the straight line OF in the horizontal plane XZ, with a velocity F= 50 ft. per sec, at an angle of 36° 52' with the axis of rotation OZ. If now, owing to gravity, the ball falls vertically while the centre moves along OF, then we must consider the plane HIH sls falling vertically with the ball. The centre moves then in a curve Oab, the projection of which upon the plane XZ is a straight line Oc. (21) Ball-players assert that the projection of this curve Oab {preceding example) upon the plane XZ is not a straight line but a curve. Explain hoiv this can be. COz Ans. We have seen in the preceding example that if the centre of the ball has a velocity Fand at the same time the ball has an angular velocity ooz CHAP. II.] RELATIVE MOTION OF A BODY. 213 around an axis OZ, tlie centre moves with the velocity Vz along the axis and at the same time the axis itself moves with the velocity Vx, Vz and Vx beinff the components of V along and perpendicular to the axis. At the same time the ball rotates about the axis OZ. The motion of is then in the straight line OV. , ° But no account has been taken of the resistance of the air. The air acts to cause a retardation of Vz and Vx . If the retardation in each case were proportional to the velocity, we should still have motion of in the straight line V. But the retardation in each case is not proportional to the velocities but more nearly proportional to the squares of the velocities. Hence the greater component is retarded propor- tionally more than the less. If then the rotation axis OZ makes an angle less than 45° with the direc- tion of V, Vz is greater than Vx and is therefore retarded proportionally more than Vx. The centre moves then in an " out-curve " OB. If, however, the axis of rotation OZ makes an angle greater than 45° with the direction of F, Fa; is greater than Vz and is therefore retarded proportionally more than Vz. The centre moves then in an " incurve." In either case the velocity is retarded least in the direction of least resistance and the centre swerves in the direction of the smallest component of V. Thus by "twisting " the ball the pitcher is able to make it curve slightly by either to right or to left according as the axis of rotation makes an angle with the velocity of projection greater or less than 45°. If the axis of rotation makes an angle of 45° with the velocity of projection, there should be no curve. If it is at right angles to the velocity of projection, there should be no curve. The cause of curvature is thus due to the resistance of the air, but it is not, as is generally supposed, due to the ball rolling upon a cushion of compressed air in front of it, since in that case we should always have curvature in one direction for one direction of rotation. In the first of our figures preceding, such action tends to increase the " out- curve." But in the second it tends to decrease the " in-curve." The " in-curve" would not be possible if this action were the only cause of curvature. It ought to be less than the out-curve, so far as this action is effective, in the figures given. If we have rotation in the opposite direction from that in the figure, or if the line representative of ojz is positive instead of negative, the rolling of the ball, if any, upon a cushion of compressed air in front of it would act to decrease the ' ' out-curve " and increase the ' ' in-curve. " Relative Motion of a Body. — When a body at any instant has two simultaneous motions we can consider the body itself as having one of these motions and the space occupied by the body as having the other. The first motion is then that which the body would appear to have to an observer in space moving with space and unaware of his own motion. We call it therefore the relative motion of the body with reference to moving space. We have thus far seen how to determine the actual motion when we have given the relative motion and the motion of space. We have now to consider the inverse problem of how to determine the relative motion when we have given the actual motion and the mo- tion of space. We can solve the problem in two ways. We can resolve the given actual motion into two component motions one of which coincides with the given motion of space. Then the other must be the relative motion required. Or we can add to the actual motion, composed of these two component motions, a third motion equal 214 EIGID SYSTEM — TKANSLATION AND KOTATION. [CHAP. II. and opposite to the given motion of space. This will counteract one of the components and leave as a result the relative motion. Relative Motion of a Body with Reference to Space Transla- tion. — If the actual motion of a body and the motion of the space occupied by it are both motions of translation, the relative motion will be one of translation also. In such case we can treat the body and the space occupied by it as points, and thus have simulv to find the relative motion of a particle with reference to the point of moving space occupied by the particle. The relative velocity of the particle is then the resultant of the actual velocity of the particle and the velocity of the point of mov- ing space occupied by the particle, taken as acting with reversed direction. If then the actual velocity of the particle is zero, the relative velocity at any instant is always equal and opposite to that of" the point of moving space occupied by the particle at that instant. Thus if the particle P de- scribes an ellipse with reference to the fixed point O at one focus, the relative velocity of O with reference to P will be always equal and opposite to the velocity of P at any in- stant, and the apparent path of O as seen from P will be a similar ellipse with P at a focus. Relative Motion of a Body with Reference to Moving Space in General. — Any motion of a body at any instant can be re- solved into a translation of any point, and a rotation about an axis through that point (page 190). If we take for this point the point of space occupied by any particle of the body, we have translation only of this point and particle, and the relative velocity is found as in the preceding Article. The relative velocity of the particle is then, as before, the result- ant of the actual velocity of the particle and the velocity of the point of moving space occupied by the particle, taken as acting with reversed direction. We obtain then the relative path by giving to the actual path the reversed motion of space. For example, let the actual velocity of a particle P be uni- form and equal to c, and its con- stant direction be in the direc- tion AB. Let the line AB be the diameter of a circular disk which rotates clockwise with constant angular velocity o about the axis at C. We obtain the path relative to the disk by supposing rotation of the actual path AB counter-clockwise. Thus at the end of any time t the particle has traversed the distance AN = ct and the has turned clockwise through the angle oot. disk The corresponding CHAP. II.] ACCELERATION OF RELATIVE MOTION. 215 jelative position P of the particle is then would occupy if the line AB were turned <30unter-clock;wise through the angle cot. Repeating the construction for successive values of t we obtain the relative path APCD. The end D corresponds to the rotation angle ooT, where T is the time of the actual motion from Ato B. It goT = Tt, the point D would coincide with A. Let r be the radius of the disk. Then from the two equations cT = 2r, and ooT=it, we have for the condition of this coincidence of D and A, oa It If the actual path AB makes an acute angle with the axis of rotation through C, the relative path lies on the surface of a cone. Acceleration of Relative Motion. — Let a particle describe the path MN with any motion, and at the same time let this path have a motion of translation. Then we can regard the first motion as relative with reference to the second, and its acceleration /i is the relative accelera- tion. Besides this relative acceleration at any instant, the particle has the acceleration /a of the motion of trans- lation at that instant. The actual acceleration of the particle is then the resultant of the two accelerations fi and fi. It is, however, different when the path MN has any motion in general, because such motion may be resolved into a motion of translation of any point of the path and a motion of rotation about an axis through that point. Take for this point the point of space occupied by the particle at any instant. Then we have besides the accelera- tion /i of the particle in its path, and the acceleration /a of the point of space occupied by the particle, a third acceleration, /s, due to the rotation, which we can determine as follows: ^ ,, X- ^ rni- • Let V be the relative velocity of the particle. Then in an in- definitely small time dt, vdt will be the element MN of the relative path This element in the time dt is translated to PQ and at the same time has the angular velocity oo about the point of space oc- ^"^Let the axis OP through this point piake the angle 6 with P^. Then at the end of the time dt, 4> will be at R. If .A is the accelera^ tion in the direction QR, then the distance QR will be QB = ^f»dt\ 216 EIGID SYSTEM — TRANSLATION AND ROTATION. [CHAP. II. The radius of rotation is OQ = vdt sin fi. Hence the distance QR is also given by QR = OQ . codt = vdt sin . oadt. Equating these two values of QR, we obtain /$ = 2vw sin Q. Hence we see that the actual acceleration of the particle in general is the resultant of three accelerations : The first, /i, is the acceleration of the relative motion of the particle. The second, fa is the acceleration of the point of space occupied by the particle. The third, fa, is equal in magnitude to twice the product of the relative velocity v of the particle, the angular velocity oo of the point of space occupied by the particle, and the sine of the angle which the element of the relative path makes with the axis through the point of space occupied by the particle. Its direction is at right angles to the plane of this axis and element, and it acts in the direction given by the rotation. If then fi,ft, fa. Fig. 1, represent these accelerations, we have by completmg the polygon in Fig. 2 the actual acceleration /. Fig. 1. Fig. 3. Inversely, if/, fa and fa are given and it is required to find the relative acceleration /i, we must take /a and /a reversed in direction. Illustrations. — Let a particle P move at any instant with the velocity v in the direction of a diameter of a circular disk. Let the disk at this instant have an angular velocity oo about its axis at C, and the distance CP of the particle from the axis be x. Then the relative acceleration is dv fi — -^j along the diameter. The dt acceleration of the point of space occupied by the particle is /a = xoo"", along the diameter. This is the central acceleration of the point P due to rotation about C. We have also /a = 2voo, acting at right angles to the plane of the element of the relative path and the axis through P parallel to the axis at C, and it acts in the direction given by the rotation as shown. The angle 9 which the element of the path makes with the parallel axis at P is 90°, and hence sin = 1. The actual acceleration is the resultant of these three accelerations. CHAP. II.] ILLUSTKATIONS. 217 Let a particle P move at any instant with the velocity v in the circumference of a circle of radius r. about the centre d, and at the same time let the centre Ci revolve about the point C» with the angular velocity go. Let the distance of P from Ca be r^. Then the relative acceleration /i is the resultant of the tangential dv v^ acceleration -:— and the central acceleration — acting towards Ci. at ri The acceleration /a of the point of space occupied by the particle is n&j' acting towards Ca. We have also /s = 2va) acting at right angles to the plane of the element of the relative path and the axis through P parallel to the axis at Ca, and it acts in the direction J^=rco8X.co ^7i=2»w«fnA given by the rotation as shown. The actual acceleration is the resultant of these three accelerations. 818 EIGID SYSTEM — TRANSLATION AND EOTATION. [CHAP. II. Let a particle P move on a meridian of the earth and have at any instant the velocity v and the tangential acceleration , . If r is the radius of the earth, the central acceleration is — and the r relative acceleration /i is the resultant of -rr and — . If A is the latitude of P, r cos A. is the radius of rotation ; and if &> is the angular velocity of the earth, the acceleration ^ of the point of space occupied by the particle is/2 = rcosA . &5^ We have also fa — 2voo sin A acting at right angles to the plane of the element of the relative path and the axis through P parallel to the earth's axis, that is, tangent to the latitude circle at P. It acts towards the east. The actual acceleration is the resultant of these three accelerations. CHAPTEK in. GENERAL ANALYTICAL RELATIONS FOR A POINT OF A RIGID ROTATING SYSTEM. EULER'S GEOMETRIC EQUATIONS. General Analytical Relations for a Point of a Rigid Rotating System. — Let a rigid system rotate at any instant about the axis IC with the angular velocity co and the angular acceleration a. Take any point O of the system as origin, and let the direction- cosines of fij be cos a, cos p, cos y. Then we have for the components of co and a cox= oa COS a, a)y= 00 COS /?, coz = co COS y ; ax= oc COS a, ay= a COS /?, «« = a COS y ; COS a — — = — , COS P = = — -, COS y — — 00 a 00 a <» and since cos'' a + cos'' /3 + cos" y = 1, (1) (2) Let (x «, 2) be the co-ordinates of any pomt on the axis IC, and <x' V z') the co-ordinates of any point P whose distance from the axis is PC = r. Then, as we have seen (page 190), we can resolve 219 220 POINT OF RIGID ROTATING SYSTEM : GENERAL RELATIONS. [CH.III. the rotation about IC into an equal angular velocity about a paral- lel axis through the origin O and a velocity u = tap of the origin O, where p is the distance 10 of the origin from the axis. The components of this velocity of O are as on page 179, equa- tion (1): Vx = oozy — 00 yZ ; Vy = OOxZ — COzX ; Vz = oOyX — ajxy. (3) The components of the linear velocity v' of P due to rotation about the parallel axis through the origin O are as on page 198^ equations (6) : v'x = ooyz — cozy' ; 1 V'y = cozx' — coxZ' ; I (4) V'z — ODxy' — OOyX. J We have thus the total components of the velocity P, just as on. page 198, equation (7): Vx=vx+ v'x ; 1 Vy=Vy+ V'y\ \ (6> Vz = Vz-\- v'z. The components of the linear tangential acceleration of P due to rotation about the parallel axis through O, we see from (4), are given by ftx = ocyZ' — azy ; fty = azX' — axZ ; ■ f tz = ocxyf — ayX . Since poo = v and 'p'^oo^ = v'^ — Vx^ + %" + r/, we have from (3) J _ (oozy — coyZf (coxZ — cazXf (ooyX — ooxVf (6) (7) Let f'n be the normal linear acceleration of the point P due to rotation about the parallel axis through the origin O. Then f'n = v'oo ; and since velocity in the hodograph is the normal acceleration in the path (page 52), we have directly from the figure, for the com- ponents otj'n, fvx = v'zWi/ — V',,a)z ; 1 fny = t^'xooi — v'zGOx :, > (^) J nz =^ V yOOx — V xOO^. J We have then for the components of the acceleration /' of the point P, from (8) and (6), fx =fnx +ftx = {V'zooy — V'ycoz) + (ayZ' — azy')] '] fy =fnv +fty = (v'xooz — v'zOOx) + {cczX — axZ')\ \ - • (9) fz =f'nz + f'tZ = {V'yax - V'xCOy) + {fiCxlf - UyZ'). J CHAP. III.] KINEMATICS OF A RIGID SYSTEM. 221 If we put for v'x, v'y, v'z their values as given in (4), we have o' — axZ') ; I fz = i.ooxX' + ooyy' + a)zZ)oo^ — aPz + {axV — ocyZf). J fx = (cJjrCC' + oOyy' + (iazZ')oox — oorx' + {cxyz' — azy); f'y = {ojxJC + GOyy' + OL)zZ')coy — oaY + {oczX' — axZ') ; I ■ • (10) Equations (10) give the valueg of the components of the linear acceleration/ of any point P of the system whose co-ordinates are (x', y', z'), in terms of these co-ordinates and the components of a> and a. The moments of the component linear accelerations with refer- ence to the origin O are fxVy' + 2f\ fy Vx" + z!\ fz\^x'^ + y\ . . . (11) For the moments about the axes of the components of rf and /' we have: about X parallel to plane YZ, Mx = tfzy'— Vyz', or fzy -fyZ' ; 1 " Y " " " ZX, My = if xz'—v'zx, or fxZ—f;^;Ui3) " Z " " " XY,Mz^v'yXf-v'xy', or fyx-fxy'.] The resultant moment in both cases is given by Mr=VMx'' + My^ + Mz'' (13) Its line representative has the direction-cosines (14) Mx My Mz My' Mr' Mr Looking along the line representative towards the origin the rota- tion is counter-clockwise. [We can deduce equations (9) directly by the Calculus. Thus if we dif- ferentiate the values of v'x, i>'y, i> z given by (4), then, since and we have at once dt='"" ■df='^- Tt-"^'' dwx dooy dooz ——- = ax, --JT- = ocy, —^ = az, dt dt dt f^ = -^ = {v'zcoy - vycoz) -f {ayz' - azjH; at fy = -^ = (i/xa>z - v'zoox) -f {azx' - aa«'); fg = ^ = {vycox - v'xGOy) -f {axy' - ayz'). Euler's Geometrical Equations.— To determine the geometrical equations beticeen the motion of a rigid system m space and th^ angular velocity of the system about an axis in the system. 222 EULER's geometrical equations. [chap. III. Let OXi, OYi, OZi be rectangular co-ordinate axes, fixed in the system and therefore rotating with it, and let the system rotate about some axis fixed in the system, and therefore making in- variable angles with these axes, so that the component angular velocities in the co-ordinate planes are (»a:, , o^j/, , «»2,. We take direction of rotation as al- ways about X from F. to Z, 1 1*^'*'!^' the op- " F. " Z, "X !>positedi- " Zi " Jf " F ' rection ' ' J negative. Let now OX, OY, OZhe rect- angular co-ordinate axes whose directions m space are invariable. For instance, the axis OZ may be always directed towards the North Pole, then XY is the plane of the celestial equator. Let the point O be taken as the centre of a sphere of radius r. Let X, F, Z and Xi , Fi , Zi be the points in which this sphere is pierced by the fixed and moving axes. Let the axes OXi , OF , OZi have the initial positions OX, OY, OZ. First turn the system about OZ as an axis through the angle XZP = rp, so that OX moves to OP, and OF to OD. Then turn the system about OD as an axis through the angle ZOZi — 6, so that OP moves to OE, and OZ to OZu Finally turn the system about OZi as an axis through the angle EZXi = 0, so that OE moves to OXi , and OD to OF. It is required to find the geometric relations between 6, cp, ^ and GOxj , 00 y., coz^ as the system rotates. These geometric relations are called Euler's Geometric Equations. Let the angular velocity of Zi perpendicular to the plane ZOZi at any instant be denoted by ~-. This is called the angular velocity of precession. Let the angular velocity of Zi along ZZi at the same do instant be denoted by . , . This is called the angular velocity of nu- tation. Let the angular velocity of Xi with reference to E at that instant be denoted by ^v. at Draw ZiN perpendicular to OZ. Then ZiN = r sin 0, and the linear velocity at any instant of Z, perpendicular to the plane ZOZ\ M do is r sin . -^, and along ZZi at the same instant it is r . The linear dt at velocity at the same instant of Zi along FiZi is ra>xi , and along ZiXi it is ro),,^. We have then directly from the figure d^ ^ , • ^ r^^ = rooy^ cos -f- 7-00x1 sin ; r sm e . -— - = rooy. sm — rco^, cos <p. at CHAP. III.] KINEMATICS OF A RIGID SYSTEM. Since the radius r cancels out, do jT = ooyi COS (p + ooxj sin (f>; dtl^ sm 9 ^ = (»j,, sin — GJx, COS (p. 223^ (1> Combining these two equations, de d^ <»a;, = ^ sin — ^ sin e cos ; ooy^ = ^ COS + -5^ sin S sm 0. (2> In the same way by drawing a perpendicular from E to OZOE we have the linear velocity of E perpendicular to ZOE equal to r cos e ^, and of Xi relative to E along EX^ , r^. etc d^ The whole velocity of X, in space along X, Yi is rcog,. Hence dip . d0 '"^i = TT* cose + -5- dt dt Equations (1), (2) and (3) are Euler's Greometric Equations. (3> EXAMPLES. (1) Deduce the angular velocities oox, coy, ooz about the fixed axiSy in terms of 6, 0, rp. Ans. Let co,- be tlie resultant angular velocity about the fixed axes. If we impress on space and also on the system, in addition to its existing motion, an angular velocity equal to — a?,- about the resultant axis of rotation, the axes OXi, OYi, OZi will become fixed and OX, OY, OZ will move with angular velocities — oox , — coy , — coz. Hence in the equations already found we have only to replace by — V. S by — 6, ^ by — 0, and coxi , <»j/, , cozj will become — cox , — ooy , — coz , and we have dO . , . d(p . . , oox= — T- sm tp + -J- sm 6 cos ip; at dt dB , , ^0 • o • I GOu = ^- COS il> 4- —rr sm sm ^ ; dt dt d(p o ■ <^^ (4) (2) Refer the axes fixed in space to the axes fixed in the system. Ans "We have simplv to Interchange in the figure Xi, Yi, Zx with X, F, Z each with each. If then the angles Q, 0, ^ are still measured as indicated in the figure, the relations connecting them with the angular velocities are ob- tained by changing cox,, coy,, coz, into — a)x, - o^y < - Mf- . ^, „ ., If we measure Q in the direction opposite to that mdicated in the fagure, the expressions for oox, ony are identical with those already found for <»ar„ ooy,. 224 EULER'S geometrical equations. [chap. III. (3) If p, q, r, are the direction-cosines of OZ with regard to the axes OXi , OFi , OZi , show that -^^ - Qoozr + rooy, = 0; dq di — rooxt + poozi dr , -- - pooy, + qoox. (5) Ans. Any one of these may be obtained by differentiating one of the ex- pressions p = — sin 6 cos 0, g' = sin sin (p, r ■= cos 9 (4) Show that the direction-cosines of either set of Euler^s axes with regard to the other are given by cos XXi = — sin tJ; sin (p -\- cos ip cos (p cos 6; ] cos YXi = cos ^ sin -|- sin tf> cos <p cos 6; V ... (6) cos ZXi = — sin cos <p. J cos XYi = — sin rp cos — cos ^ sin cos 6; \ cos YYi = cos jp cos — sin ip sin cos 6; >• . . . (7) - cos ZFi = sin 9 sin 0. j cos XZi = sin S cos ip; j cos FZi = sin 9 sin ^; V ... . (8) cos ZZi = cos 0. j Ans, We have from the figure the following spherical triangles for which we know two sides and the included angle : Triangle. Sides. Angle. _ nX =90°-4-j/» JDX:X Dx, =90-0 XDX,=B Triangle. Sides. Angle. D^y^ DX =90 + ^ Fi2>X=180-9 Z,ZX, ZZ^ = 9 Z)FF, Y'D=t I^i2>r=180-e Z2'iX,=18O-0Z,ZFi Ji2, =90 ri^i2=9O-0 ZiZ =0 Triangle. Angle. ZiPXi = 90° PFZ, FPZ. = 90 Sides. PX =il) PZ, =90-0 PZ, =90-0 PY = 90 - ^ Solving these triangles we have at once equations (6), (7), (8). (5) Prove in the same way the following : cos XiX = — sin ^ sin + cos Tp cos cos 0; \ cos FiX= — sin ip cos — cos ip sin cos 0; > cos ZiX = sin 9 cos ^. J cos Xi F = cos ^ sin + sin ^ cos cos 9 ; 1 cos Fi F = cos ip cos — sin ^ sin cos 9; > cos Zi F = sin sin ^. ) (9) (10) CHAP. III.] KINEMATICS OF A RIGID SYSTEM — ROTATION. 225 COS XiZ = — sin 6 cos <p; ^ cos YiZ = sin 6 sin 0; >■ (11) cos ZiZ = cos 6. \ (6) Mnd the relations between the co-ordinates x, y, z, of the fixed system of axes and the co-ordinates Xi, yi, Zi of the moving system. Ans. If we multiply the first of equations (6) by x, the second by y, the third by s and add, and do the same for (7) and (8), we have at once, as we see from the figure, Xi = (— sin ^ sin + cos rj} cos (p cos B)x -}- (cos rj) sin + sin ^ cos (f> cos B)y — sin 6 cos (p.z; yi = (— sin il> cos — cos ^ sin (p cos S)z }• . . (12) -|- (cos i() cos — sin V' sin cos Q)y + sin 9 sin 0.8; 2i = sin 9 cos iff.x + sin 9 sin ^.y -|-cos 9 . s. In the same way we have from equations (9), (10), (11), X = ( — sin ip sin <p + cos tfj cos <p cos Q)Xi -\- (— sin if} cos <p — cos V' sin cos 9)^i 4* sin 9 cos ^.Zi; y = (cos ^ sin 4- sin ip cos cos 9)xi }■ . (13) -|- (cos ip cos — sin ip sin cos 6)yi -)- sin 9 sin ^.0i; 2 = — sin 6 cos (p.Xi -f- sin 9 sin (p.yi -j- cos Q.Zu END OF VOLUME I. MECHANICS. INDEX. Vol. I. KINEMATICS. Absolute— motion, 13; position, 10; rest, 13. Acceleration — angular, 175; composition and resolution of, 176; concurring angular accelerations, 178; couple, 181; instantaneous axis of, 175; line representative of, 175; moment of, 177; uniform and variable, 175; unit of, 176; and translation, composition and resolution of, 189; in terms of linear, 75, 176; in terms of moment of linear, 76. Acceleration — linear, 48, 92; of gravity, 9; of relative motion, 215; in terms of angular acceleration, 75, 176; central, 86, 92, 99, 103; centre of, 92; line representative of, 49; mean and instantaneous, 48; moment of, 60; moment of, in terms of angular, 76; normal, in terms of angular, 76; proportional to force, 91; paracentric, 87; resolution and composition of, 49; resultant of, 50, 63; sign of components of, 50; tangential and normal, 52; triangle and polygon of, 49; unit of, 49; uniform and variable, 49; uniform, in- clined to direction of motion, 117. Amplitude of an oscillation, 104. Analytical relations for a rigid rotating system, 319. Angle, conical, unit of, 6; solid, 7; unit of, 5. Angular — displacement, 171; composition and resolution of, 171; concurring angular displacements, 178; couple, 181; line representative of , 170; rigid system, 170; in terms of linear, 170; and translation, composition and reso- lution of, 187. Angular revolution of a point, 71. Angular — speed, 72; mean and instantaneous, 71; numeric equations of, 72; sign of, 72; unit of, 72; in terms of linear velocity, 74; iu terms of moment of velocity, 75; in terms of normal acceleration, 76; rate of change of, 73; numeric equations of rate of change of, 73; sign of rate of change of, 73; equations of motion under different rates of change of, 73; rate of change of, in terras of linear speed, 75; rate of change of, in terms of moment of tangential acceleration, 76; graphic representation of rate of change of, 77. Angular — velocity, 174; mean and instantaneous, 174; unit of, 174; uniform and variable, 175; of a rigid system, 174; composition and resolution of, 176; in terms of linear velocity, 176; moment of, 177; concurring angular velocities, 178; resultant of, 178; couple, 181; and translation, composition and resolution of, 189; centre of parallel angular velocities, 192; result- ant, for a rigid system, 197. 227 228 INDEX. Area, measures of, 9. Axis, of earth, length of, 5, 9; of rotation, 173; of instantaneous rotation, 174; instantaneous, of angular acceleration, 175; central, 190; intersecting, ro- tation about, 195. Blackburn's pendulum, 137. Body, material, 1; projected up or down, 93; rigid, 170; relative motion of, 218. Brachistochrone, 158. Cartesian co-ordinates, 11. Centimeter, value of, 9. Central, acceleration, 86, 92, 99, 103; axis, 190. Centre, of acceleration, 92; of moments, 60; of angular velocities, 192. Circle, curvature of, 7; motion in, 76, 154, 159. Coefficient of resistance in resisting medium, 111, 113. Combined rotation and translation, 13. Components, of displacement, 36; of motion of a rigid system, 201; of veloc- ity, 44. Composition and resolution, of accelerations, 49; of displacements, 36; of an- gular displacements, 171; of angular velocities and accelerations, 176; of translation and angular displacements, 187; of translation and angular velocities or accelerations, 189; of moments, -62; of screws, 203; of veloci- ties, 43. Compound harmonic motion, 131; resolution and composition of, 131; graphic representation of, 135; application of calculus, 137. Concurring velocities, accelerations and displacements, resultant of, 37, 45, 50, 63-67; angular displacements, velocities and accelerations, 178. Configuration, 12. Conical angle, 6. Constrained motion of a point, 88, 151. Co-ordinates, Cartesian, 10. Cosines, direction, 12. Couple, angular displacement, velocity, acceleration, 181; moment of, 186. Curvature, 7; unit of, 8; of a circle, 7. Curved path, motion in, 153. Curvilinear translation. 130. Cycloid, motion in, 155; application of calculus, 157. Definition, of kinematics, 1, 15; of statics, 1; of mechanics, 1. Degree, 5. Derived unit, 2; dimensions of, 3. Differential equations of motions of a point, 81. Dimensions, of a derived unit, 3; of space, 12; of unit of speed, 15; of unit of rate of change of speed, 24. Direction cosines, 12. Displacement, .34; line representative of, 34; relative, 34, 35; triangle and poly- gon of, 35; composition and resolution of, 36; rectangular components of, 36; sign of components of, 36; resultant, 37; moment of, 60; angular, of a rigid system, 170; line representative of, 170; linear in terms of angular, 170; angular, composition and resolution of, 171; angular, concurring, 178; resultant of, 178; couple, 181; of a rigid system, 187; composition and resolution of translation and angular displacement, 187. Dyne, value of, 9. Earth's polar axis, length of, 5. Epoch, in harmonic motion, 105. Equations, homogeneous, 3, 17; of speed, 16; of rate of change of speed, 25; of motion under different rates of change of speed, 27, 71; of motion of a point under different accelerations, 50; numeric, of angular speed, 72; nu- meric, of rate of change of angular speed, 73; of motion under different rates of change of angular speed, 73; differential, of motion of a point, 81; for falling body, 93; of rotating rigid system, 176; geometrical, of Euler, 221 Euler, geometrical equations of, 321. VOL. I. — KINEMATICS. 229 Force proportional to acceleration 91 Falling body, 93. <Jeometrical equations of Euler, 221. Gram, value of, 9. Graphic representation of rate of change of soeed 29 Gravitation, law of, 99. . Gravity, acceleration of, 9, 92. Harmonic motion, simple 103; amplitude, epoch, period, phase, 105- com- ^:%^s^TSZ;^^^-' «^- '''■' ^-P^c^epre^iStioH, Hodograph, 53. Homogeneous equations, 3, 17. Inclined plane, motion on, 151. Instantaneous acceleration, 48; axis of rotation, 174; axis of angular accelera- locU 42' ^ ^^ ""^ ^^^^' ^' ^^^^^' ^^' angular speed, 72; ve- Integral curvature, 7. Intersecting axes, rotation about, 195. Invariant, 203. Isochronous oscillation, 104. Kepler's laws, 140. Kilogram, value of, 9. Kinematics, definition of, 1, 15; of a point, 91; of a rigid system, 169. Law of gravitation, 99. Laws of Kepler, 140. Length, of meridian, 5; measures of, 8; of polar axis of earth, 5; unit of 4 8- displacement, 34; velocity, 42; acceleration, 48. > » > Linear, displacement in terms of angular, 170; acceleration in terms of angu- lar, 176; velocity in terms of angular, 176. Line representative, of displacement, 34; of velocity, 43; of acceleration, 49; of angular displacement, 170; of angular velocity, 174; of angular accelera- tion, 175; of moment of displacement, velocity or acceleration, 61,- 177. Mass, measure of, 4, 9; unit of, 4, 5. Material, body, 1; particle, point, 14; system, 14. Matter, states of, 1. Mean, curvature, 7; speed, 15; rate of change of speed, 24r- Velocity, 42; accel- eration, 48; angular speed, 72; rate of change of angular speed, 73; angu- lar velocity, 174; angular acceleration, 175. Measurement, 2; unit of, 2. Measures, table of, 8; of length, 8; of area, volume, mass, 9. Mechanics, definition of, 1. Medium, resisting — motion in, 111; coefiicient of resistance in. 111, 113; mo- tion of projectiles in, 127. Meridian, length of, 5; relation of, to meter, 5. Meter, relation of, to meridian, 5; value of, 9. Moment, of displacement, velocity or acceleration, 60; line representative of, 61 ; composition and resolution of, 62; sign of, 62; centre of, 60; linear ve- locity in terms of angul \r speed, 75; of tangential acceleration in terms of rate of change of angular speed, 76; of angular velocity or acceleration, 177; of a couple, 186. Motion, 13; of translation, 13; of rotation, 13; equations of , under different rates of change of speed, 27; equations of, under different accelerations, 50; equa- tions of, under different rates of change of angular speed, 73; in a circle, 76, 154, 159; of a point, differential equations of, 81; constrained, of a point, 88, 151- rectilinear, 92; simple harmonic, 103; in resisting medium, 111; of projectiles, 117, 127; compound harmonic, 131; planetary, 139, 141; on an inclined plane, 151; in a curved path, 153; in a cycloid, 155, 157; f equations of, for rotating rigid system, 176; screw, 191; components of, for a rigid system, 201; relative, of a body, 213; relative acceleration of, 215. 230 INDEX. Newton's law of gravitation, 99. Normal, acceleration, 52. Numeric, of a quantity, 2; equations of speed, 16; equations of angular speedy 72; equations of rate of change of speed, 25; equations of rate of change of angular speed, 73. Oscillation, amplitude of, 104; isochronous, 104. Paracentric acceleration, 87. Parallel angular velocities, centre of, 192. Particle, material, 14. Path, of a point, 13; curved, motion in, 153. Pendulum, Blackburn's, 137; simple, 154, 159. Per, meaning of, 3. Period, in harmonic motion, 105. Phase, in harmonic motion, 105. Physical science, 1. Plane, polar co-ordinate, 10; inclined, motion on, 151. Planetary motion, 139; path of, 139; application of calculus, 141. Point, 10; position of, 10; of reference, 10; path of, 13; material, 14; constrained motion of, 88, 151; kinematics of, 91. Polar, axis of earth, length of, 5, 9; co-ordinates, 10; equations for motion of a. point, 83. Polygon, of displacements, 35; of velocities, 43; of accelerations, 49. Position, 10; of a point, 10. Poundal, value of, 9. Projectile, motion of, 117; motion of, in resisting medium, 187. Quantity, statement of, 2; vector, 34. Radian, 5; square, 7; solid, 7. Radius vector, 11. Range of projectiles, 119, 127. Rate of change of speed, 24; mean and instantaneous, 24; a scalar quantity, 25; dimensions of, 24; unit of, 24; numeric equations of, 25; sign of, 25; equa- tions of motion under different rates of change of, 27; graphic rejiresenta- tion of, 29. Rate of change of angular speed, 73; mean and instantaneous, 73; unit of, 72; numeric equations of, 73; sign of, 73; equations of motion under different rates of change of, 73; in terms of linear speed, 75; in terms of moment of tangential acceleration, 76; graphic representation of, 77. Rectangular co-ordinates, 11. Rectilinear motion, 92; translation and rotation combined, 193. Reference, point of, 10. Relative, displacement, 34; motion of a body, 213; acceleration of motion, 215. Relation of vertex to meridian, 5. Resistence, coefficient of, in resisting medium. 111, 113. Resisting medium, motion in, 111; ceefficient of resistance in. 111, 113; motion of projectiles in, 127. Resolution and Composition, of displacements, 36; of velocities, 43; of acceler- ations, 49; of moments, 62; of angular displacements, 171; of angular velocities and accelerations, 176; of translation and angular displacement. 187; of translation and angular velocity or acceleration, 189; of screws, 203. Rest, 12; absolute and relative, 12. Resultant, angular displacement, velocity or acceleration, 178; angular velocity and velocity of translation for a rigid system, 197. Rigid system, 12; kinematics of, 169; angular displacement of, 170; line repre- sentative of angular displacement of, 170; angular velocity of, 174; equa- tions of motion of rotating, 176; displacement of, 187; resultant angular velocity and velocity of translation of, 197; components of motion of, 201; general analytical relations of rotating, 219. Rotation, motion of, 13, 169; and translation combined, 13, 193; axis of, 173; instantaneous axis of, 174; equations of motion for, 176; condition for, 178J about intersecting axes, 195. VOL. I. — KINEMATICS. 231 ■Science, physical, 1. Screw motion, 191; composition and resolution of, 203. Sign, of speed, 16; of rate of change of speed, 25; of components of displace- ment, 36; of components of velocity, 44, of components of acceleration, 50; of components of moments, 62; of angular speed, 72; of rate of change of angular speed, 73; of angular displacement, 173; of angular velocity, 176; of angular acceleration, 176; of moment of angular velocity or acceleration, 177. Simple pendulum, 154; application of calculus, 159. Solid angle, 7; radian, 7. Space, dimensions of, 12; polar co-ordinates, 10. Standard units, 4; yard, 4; unit of length, 4; unit of mass, 5; unit of time, 4. States of matter. 1. Statement of a quantity, 2. Statics, definition of, 1. Speed, 15; mean and instantaneous, 15; dimensions of unit of, 15; numeric equa- tions of, 16; sign of, 16; a scalar quantity, 16; rate of change of, 24; mean and instantaneous rate of change of, 24; dimensions of unit of rate of change of, 24; numeric equations of rate of change of, 25; sign of rate of change of, 25; angular, 72; mean and instantaneous angular, 72; numeric equations of angular, 72; sign of angular, 72; rate of change of angular, 73; mean and instantaneous rate of change of angular, 73; numeric equa- tions of rate of change of angular, 73; sign of rate of change of angular, 73; linear in terms of angular, 74; angular in terms of moment of linear velocity, 75; angular rate of change of, in terms of linear, 75; angular in terms of normal acceleration, 76; angular rate of change of, in terms of moment of tangential acceleration, 76; graphic representation of rate of change of angular, 77. Square radian, 7. System, 12; rigid, 12; material, 14; kinematics of rigid, 169; angular displace- ment of rigid, 170; line representative of angular displacement of rigid, 170; angular velocity of rigid, 174; equations of motion of rotating rigid, 176; displacement of rigid, 187; resultant angular velocity and velocity of translation of rigid, 197; components of motion of rigid 201; general analytical relations of rigid rotating, 219. Table of measures, 8. Tangential acceleration, 52. Time, unit of, 4; standard unit of, 4. Trajectory, equation of, 118; velocity at any point of, 119; time of flight, 119; horizontal range, 119; greatest height, 120; displacement in anv direction, 120; angle of elevation, 120; envelope of trajectories. 121. Translation, motion of, 13, 91; and rotation combined, 13, 193: curved path, 130; and angular displacement, 187; and angular velocity, 189; resultant for a rigid system, 197. Triangle and polygon, of displacements, 35; of velocities, 43; of accelerations, 49. Unit, derived, 2; of measurement, 2; of length, 4, 8; of time, 4; of mass, 5, 9; of angle, 5; of conical angle, 6; of curvature, 8; of speed, 15; of rate of change of speed, 24; of velocity, 43; of acceleration, 49; of angular ve- locity, 174; of angular acceleration, 176. Vector, quantity, 34; radius, 11. j i,i ^o Velocity, 42; mean and instantaneous, 42; unit of, 43; uniform and variable, 46; line representative of, 43; resolution and composition of, 43; rectangular components of, 44; sign of components of, 44; resultant, 45, 63; moment of 60 177; moment of, in terms, of angular, 75; couple, 81; unit of angular 174; angular, in terms of linear, 176; composition and resolution of angular, 176; concurring angular, 178; composition and resolution of translation and angular, 189; centre of parallel angular, 192; resultant angular for a rigid system, 197. SHORT-TITLE CATALOGUE VV THE PUBLICATIONS OF JOHN WILEY & SONS, New York. LoNDox: C^lAP^rAX & HALL, Limited. 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